## NCERT Solutions Class 12th Maths Chapter 5: Continuity and Differentiability## Exercise 5.1
f (x) = 5x - 3 At x = 0, f (x) = f (0) = 5 (0) - 3 = -3 LHL = lim RHL = lim So, LHS = RHL = f (0) = -3. Therefore, the function f (x) is continuous at x = 0. At x = -3, f (x) = f (-3) = 5 (-3) - 3 = -15 - 3 = -18 LHL = lim RHL = lim So, LHS = RHL = f (-3) = -18. Therefore, the function f (x) is continuous at x = -3. At x = 5, f (x) = f (5) = 5 (5) - 3 = 25 - 3 = 22 LHL = lim RHL = lim So, LHS = RHL = f (5) = 22. Therefore, the function f (x) is continuous at x = 5.
f (x) = 2x At x = 3, f (x) = f (3) = 2 (3) LHL = lim RHL = lim So, LHS = RHL = f (3) = 17. Therefore, the function f (x) is continuous at x = 3.
**f (x) = x - 5****f (x) = 1/(x - 5), x ≠ 5****f (x) = (x**^{2}- 25)/(x + 5), x ≠ 5**f (x) = |x - 5|**
Let there be any real number k. At x = k, f (k) = k - 5 LHL = lim RHL = lim So, LHL = RHL = f (k) = k - 5. Therefore, the function f (x) is continuous at x = k.
Let there be any real number k such that k ≠ 5. At x = k, f (k) = 1/(k - 5) LHL = lim RHL = lim So, LHL = RHL = f (k) = 1/(k - 5). Therefore, the function f (x) is continuous at x = k.
Let there be any real number k such that k ≠ -5. At x = k, f (k) = (k LHL = lim RHL = lim So, LHL = RHL = f (k) = k - 5. Therefore, the function f (x) is continuous at x = k. Let there be any real number k. k < 5, or k = 5, or k > 5 At x = k,
f (k) = k - 5 lim So, lim Therefore, the function f is continuous for x = 5.
f (k) = 5 - k lim So, lim Therefore, the function f is continuous for x < 5.
f (k) = k - 5 lim So, lim Therefore, the function f is continuous for x > 5. Hence, the function f is continuous for all real numbers.
f (x) = x At x = n, f (n) = n lim So, lim Hence, the function f is continuous at x = n, where n is a positive integer.
At x = 0, f (0) = 0 lim So, lim Therefore, the function f is continuous at x = 0. At x = 1, f (1) = 1 LHL = lim RHL = lim So, LHL ≠ RHL. Therefore, the function f is not continuous at x = 1. At x = 2, f (2) = 5 LHL = lim RHL = lim So, LHL = RHL = f (2). Therefore, the function f is continuous at = 2.
Let there be any real number k. k < 2, or k = 2, or k > 2 At x = k,
f (k) = 2k + 3 lim So, lim Therefore, the function f is continuous for real numbers smaller than 2.
f (2) = 2k + 3 = 2(2) + 3 = 7 LHL = lim RHL = lim So, LHL ≠ RHL. Therefore, the function f is not continuous at x = 2.
f (k) = 2k - 3 lim So, lim Therefore, the function f is continuous for x > 2. Hence, the function f is discontinuous only at x = 2.
Let there be a real number k. k < -3 or k = -3 or -3 < k < 3or k = 3 or k > 3.
f (k) = -k + 3 lim So, lim Therefore, the function f is continuous for all real numbers less than -3.
f (k) = f (-3) = -(-3) + 3 = 3 + 3 = 6 LHL = lim RHL = lim So, LHL = RHL = f (k). Therefore, the function f is continuous for x = -3.
f (k) = -2k lim So, lim Therefore, the function f is continuous for all real numbers less than 3 and greater than -3.
f (k) = f (3) = 3 + 3 = 6 LHL = lim RHL = lim So, LHL ≠ RHL. Therefore, the function f is not continuous for x = 3.
f (k) = 6k + 2 lim So, lim Therefore, the function f is continuous for all real numbers greater than 3. Hence, the function f is only discontinuous at x = 3. Let there be a real number k. k < 0 or k = 0 or k > 0.
f (k) = -k/k = -1 lim So, lim Therefore, the function f is continuous for all real numbers less than 0.
f (k) = f (0) = 0 LHL = lim RHL = lim So, LHL ≠ RHL. Therefore, the function f is not continuous for x = 0.
f (k) = k/k = 1 lim So, lim Therefore, the function f is continuous for all real numbers greater than 0. Hence, the function f is only discontinuous at x = 0. Let there be a real number k. k > 1 or k = 1 or k < 1.
f (k) = k lim So, lim Therefore, the function f is continuous for all real numbers less than 1.
f (k) = f (1) = 1 + 1 = 2 LHL = lim RHL = lim So, LHL = RHL = f (k). Therefore, the function f is continuous at x = 1.
f (k) = k + 1 lim So, lim Therefore, the function f is continuous for all real numbers greater than 1. Hence, the function f is continuous for all real numbers and it has no point of discontinuity. Let there be a real number k. k > 2 or k = 2 or k < 2.
f (k) = k lim So, lim Therefore, the function f is continuous for all real numbers less than 2.
f (k) = f (2) = 2 LHL = lim RHL = lim So, LHL = RHL = f (k). Therefore, the function f is continuous at x = 2.
f (k) = k lim So, lim Therefore, the function f is continuous for all real numbers greater than 2. Hence, the function f is continuous for all real numbers and it has no point of discontinuity. Let there be a real number k. k > 1 or k = 1 or k < 1.
f (k) = k lim So, lim Therefore, the function f is continuous for all real numbers less than 1.
f (k) = f (1) = 1 LHL = lim RHL = lim So, LHL ≠ RHL. Therefore, the function f is not continuous at x = 1.
f (k) = k lim So, lim Therefore, the function f is continuous for all real numbers greater than 1. Hence, the function f is only discontinuous at x = 1.
Let there be a real number k. k > 1 or k = 1 or k < 1.
f (k) = k + 5 lim So, lim Therefore, the function f is continuous for all real numbers less than 1.
f (k) = f (1) = 1 + 5 = 6 LHL = lim RHL = lim So, LHL ≠ RHL. Therefore, the function f is not continuous at x = 1.
f (k) = k - 5 lim So, lim Therefore, the function f is continuous for all real numbers greater than 1. Hence, the function f is only discontinuous at x = 1.
Let there be a real number k. 0 ≤ k ≤ 1 or k = 1 or 1 < k < 3 or k = 3 or 3 ≤ k ≤ 10.
f (k) = 3 lim So, lim Therefore, the function f is continuous for 0 ≤ x ≤ 1.
f (k) = f (1) = 3 LHL = lim RHL = lim So, LHL ≠ RHL. Therefore, the function f is not continuous at x = 1.
f (k) = 4 lim So, lim Therefore, the function f is continuous for 1 < x < 3.
f (k) = f (3) = 5 LHL = lim RHL = lim So, LHL ≠ RHL. Therefore, the function f is not continuous at x = 3.
f (k) = 5 lim So, lim Therefore, the function f is continuous for 3 ≤ x ≤ 10. Hence, the given function f is only discontinuous at x = 1 and x = 3. Let there be a real number k. k < 0 or k = 0 or 0 ≤ k ≤ 1 or k = 1 or k > 1.
f (k) = 2k lim So, lim Therefore, the function f is continuous for x < 0.
f (k) = f (0) = 0 LHL = lim RHL = lim So, LHL = RHL = f (k). Therefore, the function f is continuous at x = 0.
f (k) = 0 lim So, lim Therefore, the function f is continuous for 0 ≤ x ≤ 1.
f (k) = f (1) = 0 LHL = lim RHL = lim So, LHL ≠ RHL. Therefore, the function f is not continuous at x = 1.
f (k) = 4k lim So, lim Therefore, the function f is continuous for x > 1. Hence, the given function f is only discontinuous at x = 1. Let there be a real number k. k < -1 or k = -1 or -1 < k ≤ 1 or k = 1 or k > 1.
f (k) = -2 lim So, lim Therefore, the function f is continuous for x < -1.
f (k) = f (-1) = -2 LHL = lim RHL = lim So, LHL = RHL = f (k). Therefore, the function f is continuous at x = -1.
f (k) = 2k lim So, lim Therefore, the function f is continuous for -1 < x ≤ 1.
f (k) = f (1) = 2 LHL = lim RHL = lim So, LHL = RHL = f (k). Therefore, the function f is continuous at x = 1.
f (k) = 2 lim So, lim Therefore, the function f is continuous for x > 1. Hence, the given function f is continuous for all real numbers.
is continuous at x = 3. Therefore, LHL = RHL = f (3). LHL = lim RHL = lim f (3) = a (3) + 1 = 3a + 1 Now, 3a + 1 = 3b + 3 = 3a + 1 3a + 1 = 3b + 3 3a = 3b + 2 a = (3b + 2)/3
Hence, the required relation between a and b is a = b + 2/3.
is continuous at x = 0. Therefore, LHL = RHL = f (0). LHL = lim RHL = lim f (0) = λ (0 Now, λ (0) = 1 = λ (0) λ = 1/0 Therefore, there is no real value of λ possible for which the given function f is continuous at x = 0. At x = 1, f (x) = f (1) = 4 + 1 = 5 LHL = lim RHL = lim So, LHL = RHL = f (k). Therefore, the function f is continuous at x = 1.
Let there be an integer k. LHL = lim RHL = lim So, LHL ≠ RHL. Hence, the function f is discontinuous for all integers.
f (π) = π lim So, lim Hence, the function f is continuous for x = π.
Let g (x) = sin x. Let there be a real number k. At x = k, g (k) = sin k LHL = lim RHL = lim So, LHL = RHL = g (k). Therefore, the function g is continuous for all real numbers. Let h (x) = cos x. Let there be a real number k. At x = k, h (k) = cos k LHL = lim RHL = lim So, LHL = RHL = h (k). Therefore, the function h is continuous for all real numbers. Now, we know that when two functions g and h are continuous then the functions g + h, g - h, and gh are all continuous as well. Hence, (a) f (x) = sin x + cos x, (b) f (x) = sin x - cos x, and (c) f (x) = sin x. cos x are all continuous functions.
Let g (x) = sin x. Let there be a real number k. At x = k, g (k) = sin k LHL = lim RHL = lim So, LHL = RHL = g (k). Therefore, the function g is continuous for all real numbers. Let h (x) = cos x. Let there be a real number k. At x = k, h (k) = cos k LHL = lim RHL = lim So, LHL = RHL = h (k). Therefore, the function h is continuous for all real numbers. Now, we know that when two functions g and h are continuous then the functions g/h where h ≠ 0, 1/h where h ≠ 0, and 1/g where g ≠ 0 are continuous as well. Thus, cosec x = 1/sin x, sin x ≠ 0 when x ≠ nπ where n ∈ Z. So, cosec x is continuous except x = nπ where n ∈ Z sec x = 1/cos x, cos x ≠ 0 when x ≠ (2n + 1)π/2 where n ∈ Z. So, sec x is continuous except x = (2n + 1)π/2 where n ∈ Z cot x = cos x/sin x, sin x ≠ 0 when x ≠ nπ where n ∈ Z. So, cot x is continuous except x = nπ where n ∈ Z
Let there be a real number k. k < 0 or k = 0 or k > 0.
f (k) = sin k/k lim So, lim Therefore, the function f is continuous for all real numbers less than 0.
f (k) = f (0) = 0 + 1 = 1 LHL = lim RHL = lim So, LHL = RHL = f (k). Therefore, the function f is continuous at x = 0.
f (k) = k + 1 lim So, lim Therefore, the function f is continuous for all real numbers greater than 0. Hence, the function f is continuous for all real numbers.
Let there be a real number k. k ≠ 0 or k = 0.
f (k) = f (0) = 0 LHL = lim We know that -1 ≤ sin 1/x ≤ 1, x ≠ 0 x lim 0 ≤ lim lim lim lim RHL = lim We know that -1 ≤ sin 1/x ≤ 1, x ≠ 0 x lim 0 ≤ lim lim lim lim So, LHL = RHL = f (k). Therefore, the function f is continuous at x = 0.
f (k) = k lim So, lim Therefore, the function f is continuous for all real numbers not equal to 0. Hence, the function f is continuous for all real numbers.
Let there be a real number k. k ≠ 0 or k = 0.
f (k) = f (0) = -1 LHL = lim RHL = lim So, LHL = RHL = f (k). Therefore, the function f is continuous at x = 0.
f (k) = sin k - cos k lim So, lim Therefore, the function f is continuous for all real numbers not equal to 0. Hence, the function f is continuous for all real numbers.
is continuous at x = π/2. Therefore, LHL = RHL = f (π/2). f (π/2) = 3 LHL = lim = lim RHL = lim = lim Now, lim lim k/2 = -k/-2 = 3 [lim k/2 = k/2 = 3
is continuous at x = 2. Therefore, LHL = RHL = f (2). f (2) = 3 LHL = lim RHL = lim Now, lim k (2 4k = 3 = 3
is continuous at x = 2. Therefore, LHL = RHL = f (π). f (π) = cos π = -1 LHL = lim RHL = lim Now, lim kπ + 1 = cos π = -1 kπ + 1 = -1 = -1 kπ = -2
is continuous at x = 5. Therefore, LHL = RHL = f (5). f (5) = 3 (5) - 5 = 15 - 5 = 10 LHL = lim RHL = lim Now, lim kπ + 1 = 3 (5) - 5 = 10 kπ + 1 = 10 = 10 kπ = 9
is continuous at x = 2 and x = 10. Therefore, LHL = RHL = f (2) and LHL = RHL = f (10)
f (2) = 5 LHL = lim RHL = lim Now, lim 5 = a (2) + b = 5 5 = 2a + b = 5
f (10) = 21 LHL = lim RHL = lim Now, lim a (10) + b = 21 = 21 10a + b = 21 = 21
So, 5 - 2a = 21 - 10a 8a = 16
AND b = 5 - 2(2) = 5 - 4
Let there be two functions g and h defined as: g (x) = cos x and h (x) = x Assuming that the functions are well defined for all real numbers, we can write the given function f as the combination of g and h as f = goh. If g and h are both continuous then goh is also continuous, i.e., f is also continuous.
Let there be a real number k. At x = k, g (k) = cos k LHL = lim RHL = lim So, LHL = RHL = g (k). Therefore, the function g is continuous for all real numbers.
Let there be a real number k. At x = k, h (k) = k lim So, lim Therefore, the function h is continuous for all real numbers. Since, g and h are both continuous functions. Thus, goh is also a continuous function. Hence, f is a continuous function.
Let there be two functions g and h defined as: g (x) = cos x and h (x) = |x|. Assuming that the functions are well defined for all real numbers, we can write the given function f as the combination of g and h as f = hog. If g and h are both continuous then hog is also continuous, i.e., f is also continuous.
Let there be a real number k. At x = k, g (k) = cos k LHL = lim RHL = lim So, LHL = RHL = g (k). Therefore, the function g is continuous for all real numbers. Now, h (x) = |x| Let there be a real number k. k < 0 or k = 0 or k > 0.
h (k) = -k lim So, lim Therefore, the function h is continuous for all real numbers less than 0.
h (k) = h (0) = 0 LHL = lim RHL = lim So, LHL = RHL = h (k). Therefore, the function h is continuous for x = 0.
h (k) = k lim So, lim Therefore, the function h is continuous for all real numbers greater than 0. Therefore, the function h is continuous for all real numbers. Since, g and h are both continuous functions. Thus, hog is also a continuous function. Hence, f is a continuous function.
Let there be two functions g and h defined as: g (x) = sin x and h (x) = |x|. Assuming that the functions are well defined for all real numbers, we can write the given function f as the combination of g and h as f = goh. If g and h are both continuous then goh is also continuous, i.e., f is also continuous.
Let there be a real number k. At x = k, g (k) = sin k LHL = lim RHL = lim So, LHL = RHL = g (k). Therefore, the function g is continuous for all real numbers. Now, h (x) = |x| Let there be a real number k. k < 0 or k = 0 or k > 0.
h (k) = -k lim So, lim Therefore, the function h is continuous for all real numbers less than 0.
h (k) = h (0) = 0 LHL = lim RHL = lim So, LHL = RHL = h (k). Therefore, the function h is continuous for x = 0.
h (k) = k lim So, lim Therefore, the function h is continuous for all real numbers greater than 0. Therefore, the function h is continuous for all real numbers. Since, g and h are both continuous functions. Thus, goh is also a continuous function. Hence, f is a continuous function.
Let there be two functions g and h defined as: g (x) = |x + 1| and h (x) = |x|. Assuming that the functions are well defined for all real numbers, we can write the given function f as the combination of g and h as f = h - g. If g and h are both continuous then h - g is also continuous, i.e., f is also continuous. g (x) = |x + 1| Let there be a real number k. k < 0 or k = 0 or k > 0.
g (k) = -(k + 1) lim So, lim Therefore, the function g is continuous for all real numbers less than 0.
g (k) = g (-1) = -1 + 1 = 0 LHL = lim RHL = lim So, LHL = RHL = g (k). Therefore, the function g is continuous for x = 0.
g (k) = k + 1 lim So, lim Therefore, the function g is continuous for all real numbers greater than 0. Therefore, the function g is continuous for all real numbers. Now, h (x) = |x| Let there be a real number k. k < 0 or k = 0 or k > 0.
h (k) = -k lim So, lim Therefore, the function h is continuous for all real numbers less than 0.
h (k) = h (0) = 0 LHL = lim RHL = lim So, LHL = RHL = h (k). Therefore, the function h is continuous for x = 0.
h (k) = k lim So, lim Therefore, the function h is continuous for all real numbers greater than 0. Therefore, the function h is continuous for all real numbers. Since, g and h are both continuous functions. Thus, h - g is also a continuous function. Hence, f is a continuous function. ## Exercise 5.2
Let y = sin (x Differentiate both sides with respect to x dy/dx = d/dx [sin (x dy/dx = (cos (x dy/dx = (cos (x dy/dx = (cos (x dy/dx = 2x cos (x
Let y = cos (sin x) Differentiate both sides with respect to x dy/dx = d/dx [cos (sin x)] dy/dx = (-sin (sin x)) d/dx (sin x) dy/dx = (-sin (sin x)) (cos x) dy/dx = -sin (sin x) cos x
Let y = sin (ax + b) Differentiate both sides with respect to x dy/dx = d/dx [sin (ax + b)] dy/dx = (cos (ax + b)) d/dx (ax + b) dy/dx = (cos (ax + b)) [a d/dx (x) + d/dx (b)] dy/dx = (cos (ax + b)) [a + 0] dy/dx = a cos (ax + b)
Let y = sec (tan (√x)) Differentiate both sides with respect to x dy/dx = d/dx [sec (tan (√x))] dy/dx = [sec (tan (√x) tan (tan (√x))] d/dx (tan (√x)) dy/dx = [sec (tan (√x) tan (tan (√x)) sec dy/dx = [sec (tan (√x) tan (tan (√x)) sec dy/dx = 1/2√x [sec (tan (√x) tan (tan (√x)) sec
Let y = sin (ax + b)/cos (cx + d) Differentiate both sides with respect to x dy/dx = d/dx [sin (ax + b)/cos (cx + d)] dy/dx = [cos (cx + d) d/dx (sin (ax + b)) - sin (ax + b) d/dx (cos (cx + d))]/[cos dy/dx = [cos (cx + d) cos (ax + b) d/dx (ax + b) + sin (ax + b) sin (cx + d) d/dx (cx + d)]/[cos dy/dx = [cos (cx + d) cos (ax + b) (a dx/dx + 0) + sin (ax + b) sin (cx + d) (c dx/dx + 0)]/[cos dy/dx = [a cos (cx + d) cos (ax + b) + c sin (ax + b) sin (cx + d)]/[cos
Let y = cos x Differentiate both sides with respect to x dy/dx = d/dx [cos x dy/dx = (cos x dy/dx = (cos x dy/dx = (2 sin x dy/dx = (2 sin x dy/dx = 10x
Let y = 2√cot (x Differentiate both sides with respect to x dy/dx = d/dx [2√cot (x dy/dx = 2 d/dx [√cot (x dy/dx = 2 (1/2 cot x dy/dx = (1/cot x dy/dx = (1/cot x dy/dx = -2x cosec
Let y = cos (√x) Differentiate both sides with respect to x dy/dx = d/dx [cos (√x)] dy/dx = (-sin (√x)) d/dx (√x) dy/dx = (-sin (√x)) (1/2√x) dy/dx = (-sin √x)/2√x
At x = 1 LHD = lim RHD = lim So, LHD RHD. Therefore, the function f (x) = |x - 1|, x R is not differentiable at x = 1.
At x = 1 LHD = lim RHD = lim So, LHD RHD. Therefore, the greatest integer function defined by f (x) = [x], 0 < x < 3 is not differentiable at x = 1. At x = 2 LHD = lim RHD = lim So, LHD RHD. Therefore, the greatest integer function defined by f (x) = [x], 0 < x < 3 is not differentiable at x = 2. Hence, proved that the greatest integer function defined by f (x) = [x], 0 < x < 3, is not differentiable at x = 1 and x = 2. ## Exercise 5.3
2x + 3y = sin x Differentiate both sides with respect to x d/dx (2x) + d/dx (3y) = d/dx (sin x) 2 d/dx (x) + 3 dy/dx = cos x 2 + 3 dy/dx = cos x 3 dy/dx = cos x - 2 dy/dx = (cos x - 2)/3
2x + 3y = sin y Differentiate both sides with respect to x d/dx (2x) + d/dx (3y) = d/dx (sin y) 2 d/dx (x) + 3 dy/dx = cos y dy/dx 2 = cos y dy/dx - 3 dy/dx 2 = (cos y - 3) dy/dx dy/dx = 2/(cos y - 3)
cos y
ax + by Differentiate both sides with respect to x d/dx (ax) + d/dx (by a d/dx (x) + b d/dy (y a + 2by dy/dx = -sin y dy/dx 2by dy/dx + sin y dy/dx = -a (2by + sin y) dy/dx = -a dy/dx = -a/(2by + sin y)
xy + y Differentiate both sides with respect to x d/dx (xy) + d/dx (y x dy/dx + y d/dx (x) + 2y dy/dx = sec x dy/dx + y + 2y dy/dx = sec x dy/dx + 2y dy/dx - dy/dx = sec (x + 2y - 1) dy/dx = sec dy/dx = (sec
x Differentiate both sides with respect to x d/dx (x 2x d/dx (x) + x dy/dx + y d/dx (x) + 2y dy/dx = 0 2x + x dy/dx + y + 2y dy/dx = 0 (x + 2y) dy/dx = -(2x + y) dy/dx = -(2x + y)/(x + 2y)
x Differentiate both sides with respect to x d/dx (x 3x 3x (x dy/dx = -(3x
sin Differentiate both sides with respect to x d/dx (sin 2 sin y d/dx (sin y) - sin xy d/dx (xy) = 0 2 sin y cos y dy/dx - sin xy [x dy/dx + y d/dx (x)] = 0 2 sin y cos y dy/dx - sin xy (x dy/dx + y) = 0 2 sin y cos y dy/dx = (x dy/dx + y) sin xy sin 2y dy/dx = (x sin xy) dy/dx + y sin xy (sin 2y - x sin xy) dy/dx = y sin xy dy/dx = (y sin xy)/(sin 2y - x sin xy)
sin Differentiate both sides with respect to x d/dx (sin 2 sin x d/dx (sin x) + 2 cos y d/dx (cos y) = 0 2 sin x cos x + 2 cos y (-sin y) dy/dx = 0 2 sin x cos x = 2 sin y cos y dy/dx sin 2x = sin 2y dy/dx dy/dx = sin 2x/sin 2y
y = sin Let x = tan θ θ = tan Now, y = sin y = sin y = 2θ y = 2 tan-1 x Differentiate both sides with respect to x dy/dx = d/dx (2 tan dy/dx = 2 d/dx (tan dy/dx = 2/(1 + x
y = tan Let x = tan θ θ = tan Now, y = tan y = tan y = 3θ y = 3 tan Differentiate both sides with respect to x dy/dx = d/dx (3 tan dy/dx = 3 d/dx (tan dy/dx = 3/(1 + x
y = cos Let x = tan θ θ = tan Now, y = cos y = cos y = 2θ y = tan Differentiate both sides with respect to x dy/dx = d/dx (3 tan dy/dx = 1/(1 + x
y = sin Let x = tan θ θ = tan Now, y = sin y = sin y = sin y = π/2 - 2θ y = π/2 - 2 tan-1 x Differentiate both sides with respect to x dy/dx = d/dx (π/2) - d/dx (2 tan dy/dx = 0 - 2 d/dx (2 tan dy/dx = 0 - 2/(1 + x dy/dx = -2/(1 + x+)
y = cos Let x = tan θ θ = tan Now, y = cos y = cos y = cos y = π/2 - 2θ y = π/2 - 2 tan Differentiate both sides with respect to x dy/dx = d/dx (π/2) - d/dx (2 tan dy/dx = 0 - 2 d/dx (2 tan dy/dx = 0 - 2/(1 + x dy/dx = -2/(1 + x
y = sin Let x = sin θ θ = sin Now, y = sin y = sin y = sin y = sin y = 2θ y = 2 sin Differentiate both sides with respect to x dy/dx = d/dx (2 sin dy/dx = 2 d/dx (sin= x) dy/dx = 2/√(1 - x
y = sec Let x = cos θ θ = cos Now, y = sec y = sec y = sec y = 2θ y = 2 cos Differentiate both sides with respect to x dy/dx = d/dx (2 cos dy/dx = 2 d/dx (cos dy/dx = -2/√(1 - x ## Exercise 5.4
Let y = e Differentiate both sides with respect to x dy/dx = d/dx (e dy/dx = [e dy/dx = (e dy/dx = e
Let y = e Differentiate both sides with respect to x dy/dx = d/dx (e dy/dx = e dy/dx = e
Let y = e Differentiate both sides with respect to x dy/dx = d/dx (e dy/dx = e dy/dx = e dy/dx = 3x
Let y = sin (tan Differentiate both sides with respect to x dy/dx = d/dx (sin (tan dy/dx = cos (tan dy/dx = cos (tan dy/dx = -e
Let y = log (cos e Differentiate both sides with respect to x dy/dx = d/dx (log (cos e dy/dx = (1/cos e dy/dx = (1/cos e dy/dx = -e dy/dx = -e
Let y = e Differentiate both sides with respect to x dy/dx = d/dx (e dy/dx = d/dx (e dy/dx = e dy/dx = e
Let y = √(e Differentiate both sides with respect to x dy/dx = d/dx (√(e dy/dx = (1/2√(e dy/dx = e dy/dx = (e dy/dx √(e
Let y = log (log x) Differentiate both sides with respect to x dy/dx = d/dx (log (log x)) dy/dx = (1/log x) d/dx (log x) dy/dx = (1/log x) (1/x) dy/dx = 1/(x log x)
Let y = cos x/log x Differentiate both sides with respect to x dy/dx = d/dx (cos x/log x) dy/dx = [log x d/dx (cos x) - cos x d/dx (log x)]/(log x) dy/dx = [-sin x log x - cos x/x]/(log x) dy/dx = (-x sin x log x - cos x)/x (log x) dy/dx = -(x sin x log x + cos x)/x (log x)
Let y = cos (log x + e Differentiate both sides with respect to x dy/dx = d/dx (cos (log x + e dy/dx = -sin (log x + e dy/dx = -sin (log x + e dy/dx = -sin (log x + e dy/dx = (1/x + e ## Exercise 5.5
Let y = cos x .cos 2x .cos 3x Take log on both sides log y = log (cos x) + log (cos 2x) + log (cos 3x) Differentiate both sides with respect to x d/dx (log y) = d/dx [log (cos x) + log (cos 2x) + log (cos 3x)] (1/y) dy/dx = d/dx (log (cos x)) + d/dx (log (cos 2x)) + d/dx (log (cos 3x)) dy/dx = y [(1/cos x) d/dx (cos x) + (1/cos 2x) d/dx (cos 2x) + (1/cos 3x) d/dx (cos 3x)] dy/dx = y [(-sin x/cos x) d/dx (x) + (-sin 2x/cos 2x) 2 d/dx (x) + (-sin 3x/cos 3x) 3 d/dx (x)] dy/dx = y [-sin x/cos x - 2 sin 2x/cos 2x - 3 sin 3x/cos 3x] dy/dx = -y [tan x + 2 tan 2x + 3 tan 3x] dy/dx = -cos x .cos 2x .cos 3x [tan x + 2 tan 2x + 3 tan 3x]
Let y = √[(x - 1) (x - 2)/(x - 3)(x - 4)(x - 5)] Take log on both sides log y = log [√[(x - 1) (x - 2)/(x - 3)(x - 4)(x - 5)]] log y = 1/2 × [log (x - 1) + log (x - 2) - log (x - 3) - log (x - 4) - log (x - 5)] Differentiate both sides with respect to x d/dx (log y) = 1/2 × d/dx [log (x - 1) + log (x - 2) - log (x - 3) - log (x - 4) - log (x - 5)] (1/y) dy/dx = 1/2 × [d/dx (log (x - 1)) + d/dx (log (x - 2)) - d/dx (log (x - 3)) - d/dx (log (x - 4)) - d/dx (log (x - 5))] dy/dx = y/2 [(1/(x - 1)) d/dx (x - 1) + (1/(x - 2)) d/dx (x - 2) - (1/(x - 3)) d/dx (x - 3) - (1/(x - 4)) d/dx (x - 4) - (1/(x - 5)) d/dx (x - 5)] dy/dx = y/2 [1/(x - 1) + 1/(x - 2) - 1/(x - 3) - (1/(x - 4)) - 1/(x - 5)] dy/dx = 1/2 × √[(x - 1) (x - 2)/(x - 3)(x - 4)(x - 5)] [1/(x - 1) + 1/(x - 2) - 1/(x - 3) - (1/(x - 4)) - 1/(x - 5)]
Let y = (log x) Take log on both sides log y = log ((log x) log y = cos x .log (log x) Differentiate both sides with respect to x d/dx (log y) = d/dx (cos x .log (log x)) (1/y) dy/dx = cos x d/dx (log (log x)) + log (log x) d/dx (cos x) dy/dx = y [cos x (1/log x) d/dx (log x) + log (log x) (-sin x)] dy/dx = y [(1/x) cos x/log x - sin x .log (log x)] dy/dx = y [cos x/x log x - sin x .log (log x)] dy/dx = (log x)
Let y = x Let x So, y = u - v dy/dx = du/dx - dv/dx Now, u = x Take log on both sides log u = log (x log u = x log x Differentiate both sides with respect to x d/dx (log u) = d/dx (x log x) (1/u) du/dx = x d/dx (log x) + log x d/dx (x) du/dx = u [x (1/x) + log x] du/dx = x AND v = 2 Take log on both sides log v = log (2 log v = sin x .log 2 Differentiate both sides with respect to x d/dx (log v) = d/dx (sin x .log 2) (1/v) dv/dx = sin x d/dx (log 2) + log 2 d/dx (sin x) dv/dx = v cos x .log 2 dv/dx = 2 Therefore, dy/dx = du/dx - dv/dx dy/dx = x
Let y = (x + 3) Take log on both sides log y = log (x + 3) log y = 2 log (x + 3) + 3 log (x + 4) + 4 log (x + 5) Differentiate both sides with respect to x d/dx (log y) = 2 d/dx (log (x + 3)) + 3 d/dx (log (x + 4)) + 4 d/dx (log (x + 5)) (1/y) dy/dx = 2/(x + 3) + 3/(x + 4) + 4/(x + 5) dy/dx = y [2(x + 4) (x + 5) + 3(x + 3) (x + 5) + 4(x + 3) (x + 4)]/(x + 3)(x + 4)(x + 5) dy/dx = y [2 (x dy/dx = y [2 (x dy/dx = y [2x dy/dx = y [9x dy/dx = (x + 3) dy/dx = (x + 3) .(x + 4)
Ley y = (x + 1/x) Let u = (x + 1/x) So, y = u + v dy/dx = du/dx + dv/dx Now, u = (x + 1/x) Take log on both sides log u = log (x + 1/x) log u = x log (x + 1/x) Differentiate both sides with respect to x d/dx (log u) = d/dx (x log (x + 1/x)) (1/u) du/dx = x d/dx (log (x + 1/x)) + log (x + 1/x) d/dx (x) du/dx = u [x/(x + 1/x) d/dx (x + 1/x) + log (x + 1/x)] du/dx = u [x/(x + 1/x) (d/dx (x) + d/dx (1/x)) + log (x + 1/x)] du/dx = u [x du/dx = u [x du/dx = (x + 1/x) AND v = x Take log on both sides log v = log (x log v = (1 + 1/x) log x Differentiate both sides with respect to x d/dx (log v) = d/dx ((1 + 1/x) log x) (1/v) dv/dx = (1 + 1/x) d/dx (log x) + log x d/dx (1 + 1/x) dv/dx = v [(1 + 1/x) (1/x) + log x (-1/x dv/dx = v [(x + 1)/x dv/dx = x Therefore, dy/dx = du/dx + dv/dx dy/dx = (x + 1/x)
Let y = (log x) Let u = (log x) So, y = u + v dy/dx = du/dx + dv/dx Now, u = (log x) Take log on both sides log u = log (log x) log u = x log (log x) Differentiate both sides with respect to x d/dx (log u) = d/dx (x log (log x)) (1/u) du/dx = x d/dx (log (log x)) + log (log x) d/dx (x) (1/u) du/dx = (x/log x) d/dx (log x) + log (log x) (1/u) du/dx = (x/log x) (1/x) + log (log x) (1/u) du/dx = (1/log x) + log (log x) du/dx = u [1/log x + log (log x)] du/dx = (log x) du/dx = (log x) AND v = x Take log on both sides log v = log (x log v = log x. log x Differentiate both sides with respect to x d/dx (log v) = d/dx (log x .log x) (1/v) dv/dx = log x d/dx (log x) + log x d/dx (log x) (1/v) dv/dx = (log x/x + (log x)/x dv/dx = v [log x + log x]/x dv/dx = x dv/dx = x Therefore, dy/dx = du/dx + dv/dx dy/dx = (log x)
Let y = (sin x) Let u = (sin x) So, y = u + v dy/dx = du/dx + dv/dx Now, u = (sin x) Take log on both sides log u = log (sin x) log u = x log (sin x) Differentiate both sides with respect to x d/dx (log u) = d/dx (x log (sin x)) (1/u) du/dx = x d/dx (log (sin x)) + log (sin x) d/dx (x) (1/u) du/dx = (x/sin x) d/dx (sin x) + log (sin x) (1/u) du/dx = (x/sin x) (cos x) + log (sin x) du/dx = u [x cot x + log (sin x)] du/dx = (sin x) AND v = sin Take log on both sides log v = log (sin Differentiate both sides with respect to x d/dx (log v) = d/dx (log (sin (1/v) dv/dx = 1/(sin (1/v) dv/dx = (1/sin (1/v) dv/dx = (1/sin dv/dx = v [(1/sin dv/dx = (sin dv/dx = (1/√(1 - x)) (1/2√x) dv/dx = (1/2√(x - x Therefore, dy/dx = du/dx + dv/dx dy/dx = (sin x)
Let y = x Let u = x So, y = u + v dy/dx = du/dx + dv/dx Now, u = x Take log on both sides log u = log (x log u = sin x .log x Differentiate both sides with respect to x d/dx (log u) = d/dx (sin x .log x) (1/u) du/dx = sin x d/dx (log x) + log x d/dx (sin x) (1/u) du/dx = sin x/x + log x .cos x du/dx = u [sin x/x + log x .cos x] du/dx = x du/dx = x AND v = (sin x) Take log on both sides log v = log (sin x) log v = cos x log (sin x) Differentiate both sides with respect to x d/dx (log v) = d/dx (cos x log (sin x)) (1/v) dv/dx = cos x d/dx (log (sin x)) + log (sin x) d/dx (cos x) (1/v) dv/dx = (cos x/sin x) d/dx (sin x) + log (sin x) .(-sin x) (1/v) dv/dx = (cot x) (cos x) - sin x .log (sin x) dv/dx = v [(cot x) (cos x) - sin x .log (sin x)] dv/dx = (sin x) Therefore, dy/dx = du/dx + dv/dx dy/dx = x
Let y = x Let u = x So, y = u + v dy/dx = du/dx + dv/dx Now, u = x Take log on both sides log u = log (x log u = x cos x .log x Differentiate both sides with respect to x d/dx (log u) = d/dx (x cos x .log x) (1/u) du/dx = x cos x d/dx (log x) + log x d/dx (x cos x) (1/u) du/dx = (x cos x)/x + log x (x d/dx (cos x) + cos x d/dx (x)) (1/u) du/dx = cos x + log x (-x sin x + cos x ) (1/u) du/dx = cos x - x sin x .log x + cos x .log x du/dx = u [cos x - x sin x .log x + cos x .log x] du/dx = x AND v = (x Take log on both sides log v = log ((x log v = log (x Differentiate both sides with respect to x d/dx (log v) = d/dx [log (x (1/v) dv/dx = (1/(x (1/v) dv/dx = 2x/(x dv/dx = v [2x/(x dv/dx = (x dv/dx = (x dv/dx = (x dv/dx = (x dv/dx = -4x/(x dv/dx = -4x/(x Therefore, dy/dx = du/dx + dv/dx dy/dx = x
Let y = (x cos x) Let u = (x cos x) So, y = u + v dy/dx = du/dx + dv/dx Now, u = (x cos x) Take log on both sides log u = log (x cos x) log u = x log (x cos x) Differentiate both sides with respect to x d/dx (log u) = d/dx (x log (x cos x)) (1/u) du/dx = x d/dx (log (x cos x)) + log (x cos x) d/dx (x) (1/u) du/dx = (x/(x cos x)) d/dx (x cos x) + log (x cos x) (1/u) du/dx = (1/cos x) (x d/dx (cos x) + cos x d/dx (x)) + log (x cos x) (1/u) du/dx = (1/cos x) (-x sin x + cos x) + log (x cos x) (1/u) du/dx = -x tan x + 1 + log (x cos x) du/dx = u [1 - x tan x + log (x cos x)] du/dx = (x cos x) AND v = (x sin x) Take log on both sides log v = log (x sin x) log v = (1/x) log (x sin x) Differentiate both sides with respect to x d/dx (log v) = d/dx (x (1/v) dv/dx = (x (1/v) dv/dx = (1/(x (1/v) dv/dx = (1/(x (1/v) dv/dx = (1/(x (1/v) dv/dx = (x cos x + sin x)/(x (1/v) dv/dx = (x cot x + 1)/x (1/v) dv/dx = (x cot x + 1 - log (x sin x))/x (1/v) dv/dx = v [(x cot x + 1 - log (x sin x))/x (1/v) dv/dx = (x sin x) Therefore, dy/dx = du/dx + dv/dx dy/dx = (x cos x)
Let y = x Let u = x So, y = u + v dy/dx = du/dx + dv/dx 0 = du/dx + dv/dx Now, u = x Take log on both sides log u = log x log u = y log x Differentiate both sides with respect to x d/dx (log u) = d/dx (y log x) (1/u) du/dx = y d/dx (log x) + log x dy/dx (1/u) du/dx = y/x + log x .dy/dx du/dx = u [y/x + log x .dy/dx] du/dx = x AND v = y Take log on both sides log v = log y log v = x log y Differentiate both sides with respect to x d/dx (log v) = d/dx (x log y) (1/v) dv/dx = x d/dx (log y) + log y d/dx (x) (1/v) dv/dx = (x/y) dy/dx + log y dv/dx = v [(x/y) dy/dx + log y] dv/dx = y Therefore, 0 = du/dx + dv/dx 0 = x 0 = (x 0 = (x 0 - (x dy/dx = -[(x
y Take log on both sides log y x log y = y log x Differentiate both sides with respect to x d/dx (x log y) = d/dx (y log x) x d/dx (log y) + log y d/dx (x) = y d/dx (log x) + log x dy/dx (x/y) dy/dx + log y = y/x + log x .dy/dx (x/y) dy/dx - log x .dy/dx = y/x - log y (x/y - log x) dy/dx = y/x - log y dy/dx = (y/x - log y)/(x/y - log x) dy/dx = [(y - x log y)/x]/[(x - y log x)/y] dy/dx = y(y - x log y)/x(x - y log x)
(cos x) Take log on both sides log (cos x) y log (cos x) = x log (cos y) Differentiate both sides with respect to x d/dx (y log (cos x)) = d/dx (x log (cos y)) y d/dx (log (cos x)) + log (cos x) dy/dx = x d/dx (log (cos y) + log (cos y) d/dx (x) (y/cos x) d/dx (cos x) + log (cos x) .dy/dx = (x/cos y) d/dx (cos y) + log (cos y) (y/cos x) (-sin x) + log (cos x) .dy/dx = (x/cos y) (-sin y) dy/dx + log (cos y) (y/cos x) (-sin x) - log (cos y) = (x/cos y) (-sin y) dy/dx - log (cos x) dy/dx -y tan x - log (cos y) = (-x tan y - log (cos x)) dy/dx dy/dx = [-y tan x - log (cos y)]/[-x tan y - log (cos x)] dy/dx = -[y tan x + log (cos y)]/-[x tan y + log (cos x)] dy/dx = [y tan x + log (cos y)]/[x tan y + log (cos x)]
Take log on both sides log xy = log e log x + log y = (x - y) log e log x + log y = (x - y) Differentiate both sides with respect to x d/dx (log x) + d/dx (log y) = d/dx (x) - dy/dx 1/x + (1/y) dy/dx = 1 - dy/dx (1/y) dy/dx + dy/dx = 1 - 1/x (1/y + 1) dy/dx = (x - 1)/x ((1 + y)/y) dy/dx = (x - 1)/x dy/dx = [(x - 1)/x]/[(1 + y)/y] dy/dx = y(x - 1)/x(y + 1)
f (x) = (1 + x)(1 + x Take log on both sides log (f (x)) = log ((1 + x)(1 + x log (f (x)) = log (1 + x) + log (1 + x (1/f (x)) f' (x) = d/dx [log (1 + x) + log (1 + x (1/f (x)) f' (x) = d/dx (log (1 + x)) + d/dx (log (1 + x (1/f (x)) f' (x) = 1/(1 + x) + (1/(1 + x (1/f (x)) f' (x) = 1/(1 + x) + 2x/(1 + x f' (x) = f (x) [1/(1 + x) + 2x/(1 + x f' (x) = (1 + x)(1 + x When x = 1 f' (1) = (1 + 1)(1 + 1 f' (1) = 2 × 2 × 2 × 2 [1/2 + 2/2 + 4/2 + 8/2] f' (1) = 16 [1/2 + 1 + 2 + 4] f' (1) = 8 + 16 + 32 + 64 f' (1) = 120
Let y = (x
Differentiate both sides with respect to x dy/dx = d/dx [(x dy/dx = (x dy/dx = (x dy/dx = (x dy/dx = 3x dy/dx = 5x
y = (x y = x y = x Differentiate both sides with respect to x dy/dx = d/dx (x dy/dx = 5x dy/dx = 5x
y = (x Take log on both sides log y = log [(x log y = log (x Differentiate both sides with respect to x d/dx (log y) = d/dx (log (x (1/y) dy/dx = d/dx (log (x (1/y) dy/dx = (1/(x (1/y) dy/dx = (1/(x (1/y) dy/dx = (2x - 5)/(x (1/y) dy/dx = [(2x - 5)(x (1/y) dy/dx = [2x (1/y) dy/dx = [5x dy/dx = y [5x dy/dx = (x dy/dx = 5x Hence, all three methods give the same answer.
Let y = u .v .w
y = u .(v .w) Differentiate both sides with respect to x dy/dx = d/dx (u .(v .w)) dy/dx = u d/dx (v .w) + v .w du/dx dy/dx = u [v dw/dx + w dv/dx] + du/dx .v .w dy/dx = du/dx .v .w + u .dv/dx .w + u .v .dw/dx Hence, proved.
y = u .v .w Take log on both sides log y = log (u .v .w) log y = log u + log v + log w Differentiate both sides with respect to x d/dx (log y) = d/dx (log u + log v + log w) (1/y) dy/dx = d/dx (log u) + d/dx (log v) + d/dx (log w) (1/y) dy/dx = (1/u) du/dx + (1/v) dv/dx + (1/w) dw/dx dy/dx = y [(1/u) du/dx + (1/v) dv/dx + (1/w) dw/dx] dy/dx = u .v .w [(1/u) du/dx + (1/v) dv/dx + (1/w) dw/dx] dy/dx = (v .w) du/dx + (u .w) dv/dx + (u .v) dw/dx dy/dx = du/dx .v .w + u .dv/dx .w + u .v .dw/dx Hence, proved. ## Exercise 5.6
x = 2at So, dx/dt = d/dt (2at dx/dt = 2a d/dt (t dx/dt = 2a (2t) dx/dt = 4at AND dy/dt = d/dt (at dy/dt = a d/dt (t dy/dt = 4at Therefore, dy/dx = (dy/dt)/(dx/dt) dy/dx = (4at dy/dx = t
x = a cos θ, y = b cos θ So, dx/dθ = d/dθ (a cos θ) dx/dθ = a d/dθ (cos θ) dx/dθ = a (-sin θ) dx/dθ = -a sin θ AND dy/dθ = d/dθ (b cos θ) dy/dθ = b d/dθ (cos θ) dy/dθ = b (-sin θ) dy/dθ = -b sin θ Therefore, dy/dx = (dy/dθ)/(dx/dθ) dy/dx = (-b sin θ)/(-a sin θ) dy/dx = b/a
x = sin t and y = cos 2t So, dx/dt = d/dt (sin t) dx/dt = cos t AND dy/dt = d/dt (cos 2t) dy/dt = (-sin 2t) 2 dt/dt dy/dt = -2 sin 2t Therefore, dy/dx = (dy/dt)/(dx/dt) dy/dx = (-2 sin 2t)/(cos t) dy/dx = (-2(2 sin t cos t))/(cos t) dy/dx = -4 sin t
x = 4t and y = 4/t So, dx/dt = d/dt (4t) dx/dt = 4 d/dt (t) dx/dt = 4 AND dy/dt = d/dt (4/t) dy/dt = 4 d/dt (t dy/dt = -4/t Therefore, dy/dx = (dy/dt)/(dx/dt) dy/dx = (-4/t dy/dx = -1/t
x = cos θ - cos 2θ, y = sin θ - sin 2θ So, dx/dθ = d/dθ (cos θ - cos 2θ) dx/dθ = d/dθ (cos θ) - d/dθ (cos 2θ) dx/dθ = (-sin θ) - (-sin 2θ) 2 d/dθ (θ) dx/dθ = -sin θ + 2 sin 2θ AND dy/dθ = d/dθ (sin θ - sin 2θ) dy/dθ = d/dθ (sin θ) - d/dθ (sin 2θ) dy/dθ = cos θ - cos 2θ .2 d/dθ (θ) dy/dθ = cos θ - 2 cos 2θ Therefore, dy/dx = (dy/dθ)/(dx/dθ) dy/dx = (cos θ - 2 cos 2θ)/( -sin θ + 2 sin 2θ)
x = a (θ - sin θ) and y = a(1 + cos θ) So, dx/dθ = d/dθ (a (θ - sin θ)) dx/dθ = a d/dθ (θ - sin θ) + (θ - sin θ) d/dx (a) dx/dθ = a (1 - cos θ) AND dy/dθ = d/dθ (a(1 + cos θ)) dy/dθ = a d/dθ (1 + cos θ) + (1 + cos θ) d/dθ (a) dy/dθ = a (-sin θ) dy/dθ = -a sin θ Therefore, dy/dx = (dy/dθ)/(dx/dθ dy/dx = (-a sin θ)/(a (1 - cos θ) dy/dx = -sin θ/(1 - cos θ) dy/dx = -(2 sin θ/2 cos θ/2)/(2 sin2 θ/2) dy/dx = -(cos θ/2)/(sin θ/2) dy/dx = -cot θ/2
x = sin So, dx/dt = d/dt (sin dx/dt = [√(cos 2t) d/dt (sin dx/dt = [√(cos 2t) (3 sin dx/dt = [3 sin dx/dt = [3 sin dx/dt = [(3 sin dx/dt = [3 sin AND dy/dt = d/dt (cos dy/dt = [√(cos 2t) d/dt (cos dy/dt = [√(cos 2t) (3 cos dy/dt = [-3 cos dy/dt = [-3 cos dy/dt = [(-3 cos dy/dt = [-3 cos Therefore, dy/dx = (dy/dt)/(dx/dt) dy/dx = [(-3 cos dy/dx = cos dy/dx = cos dy/dx = [cos dy/dx = [cos t (-3 (2 cos dy/dx = [cos t (-6 cos dy/dx = [cos t (-4 cos dy/dx = -(4 cos dy/dx = -(cos 3t)/(sin 3t) dy/dx = -cot 3t
x = a (cos t + log tan t/2), y = a sin t So, dx/dt = a d/dt (cos t + log tan t/2) dx/dt = a [d/dt (cos t) + d/dt (log tan t/2)] dx/dt = a [-sin t + (1/(tan t/2)) d/dt (tan t/2)] dx/dt = a [-sin t + (sec dx/dt = a [-sin t + (1/(sin t/2 cos t/2)) (1/2)] dx/dt = a [-sin t + 1/sin t] dx/dt = a [-sin dx/dt = a (cos AND dy/dt = a d/dt (sin t) dy/dt = a cos t Therefore, dy/dx = (dy/dt)/(dx/dt) dy/dx = (a cos t)/((a cos dy/dx = sin t/cos t dy/dx = tan t
x = a sec θ, y = b tan θ So, dx/dθ = d/dθ (a sec θ) dx/dθ = a d/dθ (sec θ) dx/dθ = a sec θ tan θ AND dy/dθ = b d/dθ (tan θ) dy/dθ = b sec Therefore, dy/dx = (dy/\dθ)/(dx/dθ) dy/dx = (b sec dy/dx = (b sec θ)/(a tan θ) dy/dx = b/(a sin θ) dy/dx = (b cosec θ)/a
x = a (cos θ + θ sin θ), y = (sin θ - θ cos θ) So, dx/dθ = a d/dθ (cos θ + θ sin θ) dx/dθ = a (-sin θ + θ d/dθ (sin θ) + sin θ d/dθ (θ)) dx/dθ = a (-sin θ + θ cos θ + sin θ) dx/dθ = aθ cos θ AND dy/dθ = a d/dθ (sin θ - θ cos θ) dy/dθ = a [cos θ - (θ d/dθ (cos θ) + cos θ d/dθ (θ))] dy/dθ = a [cos θ - (-θ sin θ + cos θ)] dy/dθ = a [cos θ + θ sin θ - cos θ] dy/dθ = aθ sin θ Therefore, dy/dx = (dy/dθ)/(dx/dθ) dy/dx = (aθ sin θ)/(aθ cos θ) dy/dx = tan θ
x = √(a So, dx/dθ = d/dx (√(a dx/dθ = (1/2√(a dx/dθ = (a dx/dθ = x dx/dθ = (x log a)/√(1 - t AND dy/dθ = d/dx (√(a dy/dθ = (1/2√( a dy/dθ = (a dy/dθ = -y dy/dθ = -(y log a)/√(1 - t Therefore, dy/dx = (dy/dθ)/(dx/dθ) dy/dx = -[(y log a)/√(1 - t dy/dx = -(y log a)/(x log a) dy/dx = -y/x Hence, proved. ## Exercise 5.7
Let y = x First order derivative = dy/dx dy/dx = d/dx (x dy/dx = 2x + 3 Second order derivative = d d d d
Let y = x First order derivative = dy/dx dy/dx = d/dx (x dy/dx = 20x Second order derivative = d d d d d
Let y = x cos x First order derivative = dy/dx dy/dx = d/dx (x cos x) dy/dx = x d/dx (cos x) + cos x d/dx (x) dy/dx = -x sin x + cos x Second order derivative = d d d d d d
Let y = log x First order derivative = dy/dx dy/dx = d/dx (log x) dy/dx = 1/x Second order derivative = d d d d
Let y = x First order derivative = dy/dx dy/dx = d/dx (x dy/dx = x dy/dx = x dy/dx = x dy/dx = x Second order derivative = d d d d d d d d
Let y = e First order derivative = dy/dx dy/dx = d/dx (e dy/dx = e dy/dx = e dy/dx = 5e Second order derivative = d d d d d d d d
Let y = e First order derivative = dy/dx dy/dx = d/dx (e dy/dx = e dy/dx = -e dy/dx = -3e Second order derivative = d d d d d d d d
Let y = tan First order derivative = dy/dx dy/dx = d/dx (tan-1 x) dy/dx = 1/(1 + x Second order derivative = d d d d d d
Let y = log (log x) First order derivative = dy/dx dy/dx = d/dx (log (log x)) dy/dx = (1/log x) d/dx (log x) dy/dx = (1/log x) (1/x) dy/dx = 1/(x log x) Second order derivative = d d d d d d d
Let y = sin (log x) First order derivative = dy/dx dy/dx = d/dx (sin (log x)) dy/dx = cos (log x) d/dx (log x) dy/dx = cos (log x) (1/x) dy/dx = (cos (log x))/x Second order derivative = d d d d d d d
y = 5 cos x - 3 sin x Differentiate both sides with respect to x dy/dx = d/dx (5 cos x - 3 sin x) dy/dx = -5 sin x - 3 cos x Now, d d d So, d d Hence, proved.
y = cos cos y = x Differentiate both sides with respect to x d/dx (cos y) = d/dx (x) -sin y dy/dx = 1 dy/dx = -1/sin y Now, d d d d d d
y = 3 cos (log x) + 4 sin (log x) y y y y y y xy y y y y y y y xy xy xy xy x x Hence, proved.
y = Ae Differentiate both sides with respect to x dy/dx = d/dx (Ae dy/dx = Ae dy/dx = mAe Differentiate both sides with respect to x d d d d d Now, LHS = d = m = m = m = -mnAe = -mn (Ae = -mn (Ae = -mn (0) = 0 = RHS Hence, proved.
y = 500e Differentiate both sides with respect to x dy/dx = d/dx (500e dy/dx = 500e dy/dx = 3500e Differentiate both sides with respect to x d d d d d d d Hence, proved.
e Differentiate both sides with respect to x d/dx (e e e e 1 = -(x + 1) dy/dx dy/dx = -1/(x + 1) Differentiate both sides with respect to x d d d d d d d Hence, proved.
y = (tan y y y y (1 + x Differentiate both sides with respect to x d/dx ((1 + x (1 + x (1 + x Hence, proved. ## Miscellaneous Exercise
Let y = (3x Differentiate both sides with respect to x dy/dx = d/dx (3x dy/dx = 9 (3x dy/dx = 9 (3x dy/dx = 9 (3x dy/dx = 27 (3x
Let y = sin Differentiate both sides with respect to x dy/dx = d/dx (sin dy/dx = 3 sin dy/dx = 3 sin dy/dx = 3 sin x cos x (sin x - 2 cos
Let y = (5x) Take log on both sides log y = log (5x) log y = 3 cos 2x log 5x Differentiate both sides with respect to x (1/y) dy/dx = d/dx (3 cos 2x log 5x) (1/y) dy/dx = 3 [cos 2x d/dx (log 5x) + log 5x d/dx (cos 2x)] (1/y) dy/dx = 3 [cos 2x (1/5x) d/dx (5x) - log 5x (sin 2x) d/dx (2x)] (1/y) dy/dx = 3 [(5 cos 2x)/5x - 2 sin 2x log 5x] (1/y) dy/dx = 3 [cos 2x - 2x sin 2x log 5x]/x dy/dx = 3y [cos 2x - 2x sin 2x log 5x]/x dy/dx = 3 (5x)
Let y = sin Differentiate both sides with respect to x dy/dx = d/dx (sin dy/dx = (1/√(1 - (x√x) dy/dx = (1/√(1 - (x√x) dy/dx = (1/√(1 - (x√x) dy/dx = (1/√(1 - (x√x) dy/dx = (1/√(1 - x dy/dx = 3√x/2√(1 - x
Let y = (cos Differentiate both sides with respect to x dy/dx = d/dx [(cos dy/dx = [√(2x + 7) d/dx (cos dy/dx = [-√(2x + 7)/√(1 - x dy/dx = [-√(2x + 7)/√(1 - x dy/dx = [-√(2x + 7)/2√(1 - x dy/dx = [-√(2x + 7)/√(4 - x dy/dx = [-(2x + 7) - (√(4 - x2)) cos dy/dx = -[2x + 7 + (√(4 - x2)) cos y = cot y = cot y = cot y = cot y = x/2 Differentiate both sides with respect to x dy/dx = d/dx (x/2) dy/dx = 1/2
Let y = (log x) Take log on both sides log y = log (log x) log y = log x .log (log x) Differentiate both sides with respect to x d/dx (log y) = d/dx (log x .log (log x)) (1/y) dy/dx = log x .d/dx (log (log x)) + log (log x) .d/dx (log x) (1/y) dy/dx = log x .(1/log x) .d/dx (log x) + log (log x) (1/x) (1/y) dy/dx = 1/x + log (log x)/x (1/y) dy/dx = (1 + log (log x))/x dy/dx = y (1 + log (log x))/x dy/dx = (log x)
Let y = cos (a cos x + b sin x) Differentiate both sides with respect to x dy/dx = d/dx (cos (a cos x + b sin x)) dy/dx = (-sin (a cos x + b sin x)) d/dx (a cos x + b sin x) dy/dx = -sin (a cos x + b sin x) (-a sin x + b cos x) dy/dx = sin (a cos x + b sin x) (a sin x - b cos x) dy/dx = sin (a
Let y = (sin x - cos x) Take log on both sides log y = log ((sin x - cos x) log y = (sin x - cos x) .log (sin x - cos x) Differentiate both sides with respect to x d/dx (log y) = d/dx ((sin x - cos x) .log (sin x - cos x)) (1/y) dy/dx = (sin x - cos x) d/dx (log (sin x - cos x)) + log (sin x - cos x) .d/dx (sin x - cos x) (1/y) dy/dx = (sin x - cos x)/(sin x - cos x) .d/dx (sin x - cos x) + (cos x + sin x) .log (sin x - cos x) (1/y) dy/dx = (cos x + sin x) + (cos x + sin x) .log (sin x - cos x) (1/y) dy/dx = (cos x + sin x) [1 + log (sin x - cos x)] dy/dx = y (cos x + sin x) [1 + log (sin x - cos x)] dy/dx = (sin x - cos x)(sin x - cos x) (cos x + sin x) [1 + log (sin x - cos x)]
Let y = x So, dy/dx = du/dx + d/dx (x dy/dx = du/dx + ax Now, u = x Take log on both sides log u = log (x log u = x log x Differentiate both sides with respect to x d/dx (log u) = d/dx (x log x) (1/u) du/dx = x d/dx (log x) + log x .d/dx (x) (1/u) du/dx = x (1/x) + log x (1/u) du/dx = 1 + log x du/dx = u [1 + log x] du/dx = x AND v = a Take log on both sides log v = log (ax) log v = x log a Differentiate both sides with respect to x d/dx (log v) = d/dx (x log a) (1/v) dv/dx = x d/dx (log a) + log a .d/dx (x) (1/v) dv/dx = x d/dx (log a) + log a .d/dx (x) (1/v) dv/dx = 0 + log a (1/v) dv/dx = log a dv/dx = v [log a] dv/dx = a Therefore, dy/dx = du/dx + ax dy/dx = x
Let y = x Let u = ^{x2}So, y = u + v dy/dx = du/dx + dv/dx Now, u = x Take log on both sides log u = log x log u = (x Differentiate both sides with respect to x d/dx (log u) = d/dx ((x (1/u) du/dx = (x (1/u) du/dx = (x (1/u) du/dx = (x du/dx = u (x du/dx = x du/dx = x AND v = (x - 3) Take log on both sides log v = log (x - 3) log v = x Differentiate both sides with respect to x d/dx (log v) = d/dx (x (1/v) dv/dx = x (1/v) dv/dx = x (1/v) dv/dx = x (1/v) dv/dx = [x (1/v) dv/dx = [x dv/dx = v [x dv/dx = (x - 3) dv/dx = (x - 3) Therefore, dy/dx = du/dx + dv/dx dy/dx = x
It is given that x = 10 (t - sin t) and y = 12 (1 - cos t) So, dx/dt = d/dt (10 (t - sin t)) dx/dt = 10 d/dt (t - sin t) + (t - sin t) d/dt (10) dx/dt = 10 (1 - d/dt (sin t)) + 0 dx/dt = 10 (1 - cos t) AND dy/dt = d/dt (12 (1 - cos t)) dy/dt = 12 d/dt (1 - cos t) + (1 - cos t) d/dt (12) dy/dt = 12 (0 - d/dt (cos t)) + 0 dy/dt = 12 (-(-sin t)) dy/dt = 12 sin t Now, dy/dx = (dy/dt)/(dx/dt) dy/dx = (12 sin t)/(10 (1 - cos t)) dy/dx = (6 sin t)/(5 (1 - cos t)) dy/dx = 6 (2 sin t/2 cos t/2)/5(2 sin dy/dx = 6 (cos t/2)/5(sin t/2) dy/dx = (6 cot t/2)/5
It is given that y = sin Differentiate both sides with respect to x dy/dx = d/dx (sin dy/dx = 1/√(1 - x dy/dx = 1/√(1 - x dy/dx = 1/√(1 - x dy/dx = 1/√(1 - x dy/dx = 1/√(1 - x dy/dx = 1/√(1 - x dy/dx = 0
It is given that x√(1 + y) + y√(1 + x) = 0 x√(1 + y) = -y√(1 + x) Square both sides (x√(1 + y)) x x x x (x - y) (x + y) + xy (x - y) = 0 (x - y) [(x + y) + xy] = 0 x + y + xy = 0 y + xy = -x y (1 + x) = -x y = -x/(1 + x) Now, differentiate both sides with respect to x dy/dx = d/dx (-x/(1 + x)) dy/dx = [(1 + x) d/dx (-x) + x d/dx (1 + x)]/(1 + x) dy/dx = [(1 + x)(-1) + x (1)]/(1 + x) dy/dx = [-1 - x + x]/(1 + x) dy/dx = -1/(1 + x) So, LHS = RHS Hence, proved.
It is given that (x - a) Differentiate both sides with respect to x d/dx [(x - a) 2 (x - a) d/dx (x - a) + 2 (y - b) d/dx (y - b) = 0 2 (x - a) (1) + 2 (y - b) dy/dx = 0 2 (y - b) dy/dx = -2 (x - a) (y - b) dy/dx = -(x - a) dy/dx = -(x - a)/(y - b) Differentiate both sides with respect to x d/dx (dy/dx) = d/dx [-(x - a)/(y - b)] d d d d d d Now, LHS = [1 + (dy/dx) = [1 + (-(x - a)/(y - b)) = [1 + (x - a) = [{(y - b) = [c = [c = c = -c -c is a constant independent of a and b. Hence, proved.
It is given that cos y = x cos (a + y) x = cos y/cos (a + y) Differentiate both sides with respect to y dx/dy = d/dy (cos y/cos (a + y)) dx/dy = [cos (a + y) .d/dy (cos y) - cos y .d/dy (cos (a + y))]/cos dx/dy = [cos (a + y) (-sin y) + sin (a + y) cos y .d/dy (a + y)]/cos dx/dy = [-sin y cos (a + y) + sin (a + y) cos y]/cos dx/dy = [sin (a + y - y)]/cos dx/dy = [sin a]/cos Therefore, dy/dx = cos So, LHS = RHS Hence, proved.
It is given that x = a (cos t + t sin t) and y = a (sin t - t cos t) So, dx/dt = d/dt (a (cos t + t sin t)) dx/dt = a d/dt (cos t) + a d/dt (t sin t) dx/dt = a (-sin t) + at d/dt (sin t) + a sin t d/dt (t) dx/dt = -a sin t + at cos t + a sin t dx/dt = at cos t AND dy/dt = d/dt (a (sin t - t cos t)) dy/dt = a d/dt (sin t) - a d/dt (t cos t) dy/dt = a cos t - at d/dt (cos t) - a cos t d/dt (t) dy/dt = a cos t - at (-sin t) - a cos t dy/dt = a cos t + at sin t - a cos t dy/dt = at sin t Therefore, dy/dx = (dy/dt)/(dx/dt) dy/dx = (at sin t)/(at cos t) dy/dx = sin t/cos t dy/dx = tan t Differentiate both sides with respect to x d/dx (dy/dx) = d/dx (tan t) d d d d d
f (x) = |x|3 f (x) can be re-written as If x >= 0, f (x) = x f' (x) = d/dx (x f' (x) = 3x f'' (x) = 3 d/dx (x f'' (x) = 6x If x < 0, f (x) = -x f' (x) = -d/dx (x f' (x) = -3x f'' (x) = -3 d/dx (x f'' (x) = -6x Therefore, f'' (x) exists for all real x.
It is given that sin (A + B) = sin A cos B + cos A sin B Differentiate both sides with respect to x d/dx (sin (A + B) = d/dx (sin A cos B + cos A sin B) cos (A + B) .d/dx (A + B) = (sin A d/dx (cos B) + cos B d/dx (sin A)) + (cos A d/dx (sin B) + sin B d/dx (cos A)) cos (A + B) .(dA/dx + dB/dx) = (-sin A sin B .dB/dx + cos B cos A .dA/dx) + (cos A cos B .dB/dx - sin B sin A .dA/dx) cos (A + B) .(dA/dx + dB/dx) = (cos A cos B - sin A sin B) .dB/dx + (cos A cos B - sin A sin B) .dA/dx cos (A + B) .(dA/dx + dB/dx) = (cos A cos B - sin A sin B) (dB/dx + dA/dx) cos (A + B) = cos A cos B - sin A sin B
There exists such functions. For example, A function given by f (x) = |x - 1| + |x - 3| is continuous for all real points but not differentiable at the two points x = 1 and x = 3.
It is given that y = Differentiate both sides with respect to x dy/dx = d/dx ( dy/dx = e)^{a cos-1 x}dy/dx = a ^{2}))dy/dx = -ay/√(1 - x Take square on both sides (dy/dx) (dy/dx) (1 - x Differentiate both sides with respect to x d/dx [(1 - x (1 - x (1 - x 2 (dy/dx) [(1 - x [(1 - x (1 - x Hence, proved. Next TopicNCERT Solutions for class 12th Maths |