Author name: Prasanna

These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2/

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.2

Ex 6.2 Class 12 Question 1.
Show that the function given by f(x) = 3x + 17 is strictly increasing on R.
Solution:
f(x) = 3x + 17
Since f(x) is a polynomial function, it is continuous and differentiable in R
∴ f'(x) = 3 > 0 for all x ∈ R
Hence f'(x) is strictly increasing on R

Exercise 6.2 Class 12 NCERT Solutions Question 2.
Show that the function given by f (x) = e^2x is strictly increasing on R.
Solution:
f (x) = e^2x
Since f(x) is an exponential function, it is continuous and differentiable in R
∴ f (x) = e^2x > 0 for all x ∈ R
Hence f(x) is strictly increasing on R.

6.2 Class 12 NCERT Solutions Question 3.
Show that the function given by f (x) = sin x is
(a) strictly increasing in \(\left( 0,\frac { \pi }{ 2 } \right) \)
(b) strictly decreasing in \(\left( \frac { \pi }{ 2 } ,\pi \right) \)
(c) neither increasing nor decreasing in (0, π)
Solution:
We have f(x) = sinx
∴ f’ (x) = cosx
(a) f’ (x) = cos x is + ve in the interval \(\left( 0,\frac { \pi }{ 2 } \right) \)
⇒ f(x) is strictly increasing on \(\left( 0,\frac { \pi }{ 2 } \right) \)

(b) f’ (x) = cos x is a -ve in the interval \(\left( \frac { \pi }{ 2 } ,\pi \right) \)
⇒ f (x) is strictly decreasing in \(\left( \frac { \pi }{ 2 } ,\pi \right) \)

(c) f’ (x) = cos x is +ve in the interval \(\left( 0,\frac { \pi }{ 2 } \right) \)
while f’ (x) is -ve in the interval \(\left( \frac { \pi }{ 2 } ,\pi \right) \)
∴ f(x) is neither increasing nor decreasing in (0, π)

Ex 6.2 Class 12 Ncert Solutions Question 4.
Find the intervals in which the function f given by f(x) = 2x² – 3x is
(a) strictly increasing
(b) strictly decreasing
Solution:
f(x) = 2x² – 3x
⇒ f’ (x) = 4x – 3
⇒ f’ (x) = 0 at x = \(\frac { 3 }{ 4 }\)
The point \(x=\frac { 3 }{ 4 }\) divides the real

(i) f(x) is strictly increasing in (\(\frac { 3 }{ 4 }\), ∞)
(ii) f(x) is strictly increasing in (- ∞, \(\frac { 3 }{ 4 }\))

Ex6.2 Class 12 Question 5.
Find the intervals in which the function f given by f (x) = 2x³ – 3x² – 36x + 7 is
(a) strictly increasing
(b) strictly decreasing
Solution:
f (x) = 2x³ – 3x² – 36x + 7
Differentiating w.r.t. x,
f (x) = 6 (x – 3) (x + 2)
⇒ f’ (x) = 0 at x = 3 and x = – 2
The points x = 3, x = – 2, divide the real line into three disjoint intervals viz. (-∞, -2), (-2, 3), (3, ∞)
Now f’ (x) is +ve in the intervals (-∞, -2) and (3, ∞). Since in the interval (-∞, -2) each factor x – 3, x + 2 is -ve.
⇒ f’ (x) = + ve.
(a) f is strictly increasing in (-∞, -2)∪(3, ∞)

(b) In the interval (-2, 3), x+2 is +ve and x-3 is -ve.
f (x) = 6(x – 3)(x + 2) = + x – = -ve
∴ f is strictly decreasing in the interval (-2, 3).

Exercise 6.2 Class 12 Maths Ncert Solutions Question 6.
Find the intervals in which the following functions are strictly increasing or decreasing:
(a) x² + 2x – 5
(b) 10 – 6x – 2x²
(c) – 2x³ – 9x² – 12x + 1
(d) 6 – 9x – x²
(e) (x + 1)³(x – 3)³
Solution:
(a) Let f(x) = x² + 2x – 5
Differentiating w.r.t. x,
f'(x) = 2x + 2 = 2(x + 1)
f'(x) = 0 ⇒ 2(x + 1) = 0 = – 1
x = – 1 divides R into disjoint open intervals as (-∞, -1) and (-1, ∞)

∴ f is strictly decreasing in (-∞, -1)
f is strictly increasing in (-1, ∞)

(b) Let f(x) = 10 – 6x – 2x²
Differentiating w.r.t. x,
f'(x) = – 6 – 4x = – 2(2x + 3)
f'(x) = 0 ⇒ – 2(2x + 3) = 0

∴ f is strictly decreasing in (-∞, \(\frac { -3 }{ 2 }\)) and
f is strictly increasing in (\(\frac { 3 }{ 2 }\), ∞)

(c) Let f(x) = – 2x³ – 9x² – 12x + 1
∴ f’ (x) = – 6x² – 18x – 12
= – 6(x² + 3x + 2)
f'(x) = – 6(x + 1)(x + 2), f’ (x) = 0 gives x = -1 or x = -2
The points x = – 2 and x = – 1 divide the real line into three disjoint intervals namely ( – ∞, – 2) ( – 2, – 1) and( – 1, ∞).

f (x) is increasing in (-2, -1)
In the interval (-1, ∞) i.e.,-1 < x < ∞,(x + 1) and (x + 2) are both positive. f’ (x) = (-) (+) (+) = -ve.
⇒ f (x) is decreasing in (-1, ∞)
Hence, f (x) is increasing for – 2 < x < – 1 and decreasing for x < – 2 and x > – 1.

(d) Let f(x) = 6 – 9x – x²
Differentiating w.r.t. x,

(e) Let f(x) = (x + 1)³(x – 3)³
Differentiating w.r.t. x,
f'(x) = 3 (x² – 2x – 3)² (2x – 2)
= 6 [(x + 1) (x – 3)]² (x – 1)
= 6 (x + 1)² (x – 1 ) (x – 3)²
f'(x) = 0 ⇒ x = – 1, 1 or 3
x = – 1, x = 1 or x = 3 divide R into disjoint open intervals (-∞, -1), (-1, 1), (1, 3) and (-3, – ∞)

Thus f is strictly decreasing in (- ∞, – 1) and (- 1, 1) and f is strictly increasing in (1, 3) and (3, ∞)

Exercise 6.2 Class 12 Maths Question 7.
Show that y = log(1 + x) – \(\frac { 2x }{ 2 + x }\), x > – 1 is an increasing function of x throughout its domain.
Solution:

When x > – 1, we get y’ = (+)(+) = (+) i.e.,
y’ is positive for x > – 1
∴ When x > – 1, y = log(1 + x) – \(\frac { 2x }{ 2 + x }\) an increasing function.

Class 12 Maths Ex 6.2 NCERT Solutions Question 8.
Find the values of x for which y = [x (x – 2)]² is an increasing function.
Solution:
y = x⁴ – 4x³ + 4x²
∴ \(\frac { dy }{ dx }\) = 4x³ – 12x² + 8x
For the function to be increasing \(\frac { dy }{ dx }\) >0
4x³ – 12x² + 8x>0
⇒ 4x(x – 1)(x – 2)>0

Thus, the function is increasing for 0 < x < 1 and x > 2.

Ex 6.2 Class 12 Maths Ncert Solutions Question 9.
Prove that y = \(\frac { 4sin\theta }{ (2+cos\theta ) } -\theta \) is an increasing function of θ in \(\left[ 0,\frac { \pi }{ 2 } \right]\)
Solution:

∴ \(\frac { dy }{ dθ }\) > 0 since (4 – cos θ) is positive.
Hence y is strictly increasing on (0, \(\frac { π }{ 2 }\)).
Also y is continuous at θ = 0 and θ = \(\frac { π }{ 2 }\).
Hence y is increasing on [0, \(\frac { π }{ 2 }\)]

Class 12 Maths Chapter 6 Exercise 6.2 Question 10.
Prove that the logarithmic function is strictly increasing on (0, ∞).
Solution:
Let f (x) = log x.
The domain of f is (0, ∞).
Now, f’ (x) = \(\frac { 1 }{ x }\)
When takes the values x > 0
\(\frac { 1 }{ x }\) > 0, when x > 0,
∵ f’ (x) > 0
Hence, f (x) is an increasing function for x > 0 i.e

Another method:
From the graph, we can observe that as x increases f(x) also increases. Hence logx is an increasing function on (0, ∞).

Ex 6.2 Class 12 Maths  Question 11.
Prove that the function f given by f (x) = x² – x + 1 is neither strictly increasing nor strictly decreasing on (- 1, 1).
Solution:
let f (x) = x² – x + 1
Differentiating both sides w.r.t. x,
f'(x) = 2x – 1
f'(x) = 2(x – \(\frac { 1 }{ 2 }\)) … (1)
f'(x) = 0 ⇒ 2(x – \(\frac { 1 }{ 2 }\)) = 0 ⇒ x = \(\frac { 1 }{ 2 }\)
∴ The domain of f(x) is (-1, 1)

∴ f(x) is strictly increasing in (\(\frac { 1 }{ 2 }\), 1) and strictly decreasing in (- 1, \(\frac { 1 }{ 2 }\))
In the interval (- 1, 1) the given function is neither strictly increasing nor strictly decreasing.

Class 12 Ex 6.2 NCERT Solutions Question 12.
Which of the following functions are strictly decreasing on \(\left[ 0,\frac { \pi }{ 2 } \right] \)
i. cos x
ii. cos 2x
iii. cos 3x
iv. tan x
Solution:

sin 3x can be positive or negative.
∴ f'(x) can be positive or negative.
Hence cos 3x is neither increasing nor decreasing on (0, (\(\frac { π }{ 2 }\)).

iv. Let f(x) = tan x, x ∈ (0, \(\frac { π }{ 2 }\))
Differentiating w.r.t. x,
f'(x) = sec²x is positive
tan x is strictly increasing on (0, \(\frac { π }{ 2 }\))
Thus cos x and cos 2x are strictly decreasing on (0, \(\frac { π }{ 2 }\))

6.2 Class 12 Maths Question 13.
On which of the following intervals is the function f given by f (x) = x¹⁰⁰ + sin x – 1 strictly decreasing ?
i. (0, 1)
ii. \(\left[ \frac { \pi }{ 2 } ,\pi \right] \)
iii. \(\left[ 0,\frac { \pi }{ 2 } \right] \)
iv. none of these
Solution:
f(x) = x¹⁰⁰ + sin x – 1
Differentiating w.r.t x,
∴ f’ (x)= 100x⁹⁹ + cos x

i. When x ∈ (0, 1), x⁹⁹ > 0 and cosx > 0
∴ f'(x) is positive
i.e.,/(x) is strictly increasing on (0, 1)

ii. When x ∈ (\(\frac { π }{ 2 }\), π), 100x⁹⁹ > 100 and -1 < cosx < 0
f'(x)= 100x⁹⁹ + cosx > 100 + a negative number greater than – 1.
f'(x) is positive
∴ Hence f(x) is strictly increasing on (\(\frac { π }{ 2 }\), π)

iii. When x ∈ (0, \(\frac { π }{ 2 }\)), 100x⁹⁹ > 0 and cosx > 0
Hence f'(x) is positive
f(x) is strictly increasing on (0, \(\frac { π }{ 2 }\))
∴ f(x) is not strictly decreasing in any of the intervals (0, 1), (\(\frac { π }{ 2 }\), π) and (0, \(\frac { π }{ 2 }\)).

6.2 Maths Class 12 NCERT Solutions Question 14.
Find the least value of a such that the function f given by f (x) = x² + ax + 1 is strictly increasing on (1, 2).
Solution:
We have f (x) = x² + ax + 1
∴ f’ (x) = 2x + a.
Since f (x) is an increasing function on (1, 2)
f’ (x) > 0 for all 1 < x < 2 Now, f” (x) = 2 for all x ∈ (1, 2) ⇒ f” (x) > 0 for all x ∈ (1, 2)
⇒ f’ (x) is an increasing function on (1, 2)
⇒ f’ (x) is the least value of f’ (x) on (1, 2)
But f’ (x)>0 ∀ x∈ (1, 2)
∴ f’ (1)>0 ⇒ 2 + a > 0 ⇒ a > – 2
∴ Thus, the least value of ‘a’ is – 2.

Class 12 Maths Exercise 6.2 Question 15.
Let I be any interval disjoint from (- 1, 1). Prove that the function f given by f(x) = x + \(\frac { 1 }{ x } \) is strictly increasing on I.
Solution:
f(x) = x + \(\frac { 1 }{ x } \) and I = R – (-1, 1)
Differentiating w.r.t. x,
f'(x) = 1 – \(\frac{1}{x^{2}}\) = \(\frac{x^{2}-1}{x^{2}}\)
When x² > 1, f'(x) > 0
When x ∈ R – (- 1, 1), x² > 1
Hence f'(x) is positive.
i.e., when x ∈ 1, f(x) is strictly increasing

Question 16.
Prove that the function f given by f (x) = log sin x is strictly increasing on (0, \(\frac { π }{ 2 }\)) and strictly decreasing on (\(\frac { π }{ 2 }\), π)
Solution:
f'(x) = \(\frac { 1 }{ sin\quad x }\)cos x cot x
when 0 < x < \(\frac { \pi }{ 2 }\), f’ (x) is +ve; i.e., increasing
When \(\frac { \pi }{ 2 }\) < x < π, f’ (x) is – ve; i.e., decreasing,
∴ f (x) is decreasing. Hence, f is increasing on (0, π/2) and strictly decreasing on (π/2, π).

Question 17.
Prove that the function f given by f(x) = log cos x is strictly decreasing on (0, \(\frac { π }{ 2 }\))and strictly increasing on (\(\frac { π }{ 2 }\), π)
Solution:
Given f(x) = log cos x
f'(x) = \(\frac { 1 }{ cosx }\) (-sinx) = – tanx
In the interval \(\left( 0,\frac { \pi }{ 2 } \right)\), f’ (x) = -ve
∴ f is strictly decreasing.
In the interval \(\left( \frac { \pi }{ 2 } ,\pi \right)\), f’ (x) is + ve.
∴ f is strictly increasing in the interval.

Question 18.
Prove that the function given by
f (x) = x³ – 3x² + 3x -100 is increasing in R.
Solution:
f’ (x) = 3x² – 6x + 3
= 3 (x² – 2x + 1)
= 3 (x -1 )²
Now x ∈ R, f'(x) = (x – 1)² ≥ 0
i.e. f'(x) ≥ 0 ∀ x ∈ R; hence, f(x) is increasing on R.

Question 19.
The interval in which y = x² e^-x is increasing is
(a) (-∞, ∞)
(b) (-2 0)
(c) (2, ∞)
(d) (0, 2)
Solution:
y = x² e^-x
Differentiating w.r.t. x, we get

x = 0 and x = 2 divide the domain R into disjoint open intervals as (-∞, 0), (0, 2) and (2, ∞).
∴ f is strictly increasing on (0, 2)

These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-6-ex-6-1/

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.1

Question 1.
Find the rate of change of the area of a circle with respect to its radius r when (a) r = 3 cm (b) r = 4 cm
Solution:
Let A be the area of the circle with radius r.
A = πr²
Differentiating w.r.t. r, \(\frac { dA }{ dr }\) = 2πr
i. r = 3cm, \(\frac { dA }{ dr }\) = 2π(3) = 6πcm²/cm
ii. r = 4cm, \(\frac { dA }{ dr }\) = 2π(4) = 8πcm² /cm

Question 2.
The volume of a cube is increasing at the rate of 8 cm³/s. How fast is the surface area increasing when the length of an edge is 12 cm?
Solution:
Let x be the length of the cube, V the volume and S the surface area at time l
∴ V = x³ and S = 6x².
\(\frac { dV }{ dt }\) = 8cm³/s
V = x³
Differentiating both sides, w.r.t. t

When the edge of the cube is 12 cm, area increases at the rate of \(\frac { 8 }{ 3 }\) cm²/s.

Question 3.
The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.
Solution:
Let r be the radius of the circle.
Area of circle = πr² = A
also \(\frac { dr }{ dt }\) = 3 cm/ sec.
r = 10 m
Differentiating A w.r.t. t, we get
\(\frac { dA }{ dt }\) = 2πr.\(\frac { dr }{ dt }\)
= 2π(10)(3) = 60π cm²/sec

Question 4.
An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?
Solution:
Let V be the volume and x be the edge of the cube at time ‘t’.
V = x³
Given \(\frac { dx }{ dt }\) = 3 cm/s
Differentiating both sides, w.r.t. t
\(\frac { dV }{ dt }\) = 3x².\(\frac { dx }{ dt }\) = 3x²(3) = 9x²
When x = 10 cm, \(\frac { dV }{ dt }\) = 9 x (10)² = 900 cm³/s

Question 5.
A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?
Solution:
Let r be the radius of a wave circle:
\(\frac { dx }{ dt }\) = 5cm/sec.
A = πr²
Differentiating w.r.t. t
\(\frac { dA }{ dt }\) = 2πr\(\frac { dr }{ dt }\) = 2πr(5) = 10πr
When r = 8, \(\frac { dA }{ dt }\) = 10π(8) = 80π cm²/s

Question 6.
The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?
Solution:
The rate of change of circle w.r.t time t is given to be 0.7 cm/sec. i.e. \(\frac { dr }{ dt }\) = 0.7 cm/sec.
Now, circumference of the circle is c = 2πr
\(\frac{d \mathrm{C}}{d t}=2 \pi \cdot \frac{d r}{d t}\) = 2π(0.7) = 1.4πcm/s
The rate of increase of circumference is 1.4 π cm/s.

Question 7.
The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of
(a) the perimeter, and
(b) the area of the rectangle.
Solution:
The length x of a rectangle is decreasing at dx the rate of 5cm/min. ⇒ \(\frac { dx }{ dt }\) = – 5cm min … (i)
The width y is increasing at the rate of 4cm/min.
\(\frac { dy }{ dt }\) = 4 cm/minute
(a) p = 2x + 2y
Differentiating both sides w.r.t., t,
\(\frac { dP }{ dt }\) = 2\(\frac { dx }{ dt }\) + 2\(\frac { dy }{ dt }\)
= 2(-5) + 2(4) = – 2cm/minute
Perimeter is decreasing at the rate of 2cm/ minute

(b) A = xy
Differentiating both sides w.r.t., t,
\(\frac { dA }{ dt }\) = x.\(\frac { dy }{ dt }\) + y.\(\frac { dx }{ dt }\)
When x = 8 cm and y = 6 cm
= 32 – 30 = 2 cm²/minute
Area is increasing at the rate of 2 cm²/minute

Question 8.
A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.
Solution:
Let V be the volume and r be the radius of the balloon at time t.
Given \(\frac { dV }{ dt }\) = 900
V = \(\frac { 4 }{ 3 }\) πr³
Differentiating both sides w.r.t. t
\(\frac { dV }{ dt }\) = \(\frac { d }{ dt }\)(\(\frac { 4 }{ 3 }\) πr³) = \(\frac { d }{ dr }\)(\(\frac { 4 }{ 3 }\) πr³) \(\frac { d }{ dr }\)
900 = 4πr².\(\frac { dr }{ dt }\)
\(\frac { dr }{ dt }\) = \(\frac{900}{4 \pi r^{2}}\) =\(\frac{225}{\pi r^{2}}\)
when r = 15 cm
\(\frac { dr }{ dt }\) = \(\frac{225}{\pi(15)^{2}}\) = \(\frac{1}{\pi}\) cm/sec.
When radius is 15 cm, the rate at which the radius increases is \(\frac { 1 }{ π }\) cm/sec.

Question 9.
A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.
Solution:
Let r be the variable radius of the balloon which is in the form of sphere Vol. of the sphere
Differentiating w.r.t. r, \(\frac { dV }{ dr }\) = 4π²
When r = 10, \(\frac { dV }{ dr }\) = 4π(10)² = 400πcm³/cm
i.e., The rate at which volume of the bal¬loon is increasing with respect to the radius is 400πcm³/cm

Question 10.
A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4m away from die wall ?
Solution:

Let AB be the ladder and OB be the wall. At an instant,
let OA = x, OB = y,
x² + y² = 25 …(i)
Differentiating both sides w.r.t., t,
2x.\(\frac { dx }{ dt }\) + 2y.\(\frac { dy }{ dt }\) = 0 … (1)
x² + y² = 25 ⇒ y = \(\sqrt{25-x^{2}}\)
When x = 4m, y = \(\sqrt{25-16}\) = 3m
When x = 400 cm, y = 300 cm
(1) ⇒ (2 x 400 x 2) + (2 x 300 x \(\frac { dy }{ dt }\)) = 0
1600 + 600\(\frac { dy }{ dt }\) = 0
∴\(\frac { dy }{ dt }\) = \(\frac { -16 }{ 6 }\) = \(\frac { -8 }{ 3 }\) cm/s
∴ Height of the ladder on the wall is de-creasing at the rate of \(\frac { 8 }{ 3 }\) cm/s.

Question 11.
A particle moves along the curve 6y = x³ + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.
Solution:
We have
6y = x³ + 2 … (i)
Differentiating both sides w.r.t., t,

Hence the required points on the curve are (4, 11) and (- 4, \(\frac { -31 }{ 3 }\))

Question 12.
The radius of an air bubble is increasing at the rate of \(\frac { 1 }{ 2 }\) cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?
Solution:
Let r be the radius then V = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }\)
\(\frac { dr }{ dt } =\frac { 1 }{ 2 } \)cm/sec
\(\frac { dv }{ dt } =\frac { d }{ dt } \left( \frac { 4 }{ 3 } \pi { r }^{ 3 } \right) =\frac { 4 }{ 3 } { \pi .3r }^{ 2 }.\frac { dr }{ dt } \) = 2πr²
Hence, the rate of increase of volume when radius is 1 cm = 2π x 1² = 2π cm³/sec.

Question 13.
A balloon, which always remains spherical, has a variable diameter \(\frac { 3 }{ 2 } \)(2x+1). Find the rate of change of its volume with respect to x.
Solution:
Let r be the radius and V be the volume at time t.

Question 14.
Sand is pouring from a pipe at the rate of 12 cm³/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?
Solution:
Let V be the volume, r be the base radius and h be the height of the sand cone at time t.

When h = 4 cm, \(\frac { dh }{ dt }\) = \(\frac { 1 }{ 48π }\) cm/sec
When the height is 4 cm, the rate at which the height of the sand cone increases is \(\frac { 1 }{ 48π }\) cm/sec.

Question 15.
The total cost C (x) in Rupees associated with the production of x units of an item is given by C (x) = 0.007 x³ – 0.003 x² + 15x + 4000. Find the marginal cost when 17 units are produced.
Solution:
C(x) = 0.007x³ – 0.003x² + 15x + 4000
Marginal cost, MC = \(\frac { d }{ dx }\) C(x)
= 3(0.007)x² – 2(0.003)x + 15
= 0.021 x² – 0.006x + 15
When x = 17
MC = 0.021(17²) – (0.006)17 + 15
= 6.069 – 0.102 + 15 = 20.967
When 17 units are produced,
Marginal cost = ₹ 20.967
∴ The required marginal cost ₹ 30.02 (approx)

Question 16.
The total revenue in rupees received from the sale of JC units of a product is given by
R (x) = 13x² + 26x + 15.
Find the marginal revenue when x = 7.
Solution:
Marginal Revenue (MR)
= Rate of change of total revenue w.r.t. the
number of items sold at an instant = \(\frac { dR }{ dx } \)
We know R(x) = 13x² + 26x + 15,
MR = \(\frac { dR }{ dx } \) = 26x + 26 = 26(x+1)
Now x = 7, MR = 26 (x + 1) = 26 (7 + 1) = 208
Hence, Marginal Revenue = Rs 208.

Question 17.
The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
(a) 10π
(b) 12π
(c) 8π
(d) 11π
Solution:
(b) ∵ A = πr² ⇒ \(\frac { dA }{ dr } \) = 2π x 6 = 12π cm²/radius

Question 18.
The total revenue in Rupees received from the sale of x units of a product is given by R (x) = 3x² + 36x + 5. The marginal revenue, when x = 15 is
(a) 116
(b) 96
(c) 90
(d) 126
Solution:
(d) R(x) = 3x² + 36x + 5 ,
MR = \(\frac { dR }{ dx } \) = 6x + 36 ,
At x = 15; \(\frac { dR }{ dx } \) = 6 x 15 + 36 = 90 + 36 = 126

These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-5-ex-5-8/

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.8

Ex 5.7 Class 12 Maths Ncert Solutions Question 1.
Verify Rolle’s theorem for the function
f(x) = x² + 2x – 8, x ∈ [- 4, 2]
Solution:
Now f(x) = x² + 2x – 8 is a polynomial
∴ It is continuous and derivable in its domain x ∈ R.
Hence it is continuous in the interval [- 4, 2] and derivable in the interval (- 4, 2)
f(-4) = (- 4)² + 2(- 4) – 8 = 16 – 8 – 8 = 0,
f(2) = 2² + 4 – 8 = 8 – 8 = 0
Conditions of Rolle’s theorem are satisfied.
f'(x) = 2x + 2
∴ f’ (c) = 2c + 2 = 0
or c = – 1, c = – 1 ∈ [- 4, 2]
Thus f’ (c) = 0 at c = – 1.

Ex 5.7 Class 12 NCERT Solutions Question 2.
Examine if Rolle’s theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle’s theorem from these example?
(i) f(x) = [x] for x ∈ [5, 9]
(ii) f (x) = [x] for x ∈ [-2, 2]
(iii) f (x) = x² – 1 for x ∈ [1, 2]
Solution:
(i) In the interval [5, 9], f (x) = [x] is neither continuous nor derivable at x = 6, 7, 8 Hence Rolle’s theorem is not applicable
(ii) f (x) = [x] is not continuous and derivable at – 1, 0, 1. Hence Rolle’s theorem is not applicable.
(iii) f(x) = (x² – 1), f(1) = 1 – 1 = 0,
f(2) = 22 – 1 = 3
f(a) ≠ f(b)
Though it is continous and derivable in the interval [1,2].
Rolle’s theorem is not applicable.
In case of converse if f (c) = 0, c ∈ [a, b] then conditions of rolle’s theorem are not true.
(i) f (x) = [x] is the greatest integer less than or equal to x.
∴ f(x) = 0, But fis neither continuous nor differentiable in the interval [5, 9].

(ii) Here also, theough f (x) = 0, but f is neither continuous nor differentiable in the interval [- 2, 2].

(iii) f (x) = x² – 1, f'(x) = 2x. Here f'(x) is not zero in the [1, 2], So f (2) ≠ f’ (2).

Question 3.
If f: [- 5, 5] → R is a differentiable function and if f (x) does not vanish anywhere then prove that f (- 5) ≠ f (5).
Solution:
For Rolle’s theorem
If (i) f is continuous in [a, b]
(ii) f is derivable in [a, b]
(iii) f (a) = f (b)
then f’ (c) = 0, c ∈ (a, b)
∴ f is continuous and derivable
i.e., f'(c) ≠ 0. Hence \(\frac{f(5)-f(-5)}{10}\)
but f (c) ≠ 0 ⇒ f(a) ≠ f(b)
⇒ f(-5) ≠ f(5)

Question 4.
Verify Mean Value Theorem, if
f (x) = x² – 4x – 3 in the interval [a, b], where a = 1 and b = 4.
Solution:
f (x) = x² – 4x – 3. It being a polynomial it is continuous in the interval [1, 4] and derivable in (1,4), So all the condition of mean value theorem hold.
then f’ (x) = 2x – 4,
f'(c) = 2c – 4
f(4) = 16 – 16 – 3 = – 3,
f(1) = 1 – 4 – 3 = – 6
∴ f'(c) = 0 \(\frac{f(b)-f(a)}{b-a}\) = \(\frac{f(4)-f(1)}{4-1}\)
⇒ 2c – 4 = \(\frac{-3-6}{4-1}\)
⇒ 2c – 4 = 1 ⇒ c = \(\frac{5}{2}\) ∈ (1, 4)
∴ Mean Value Theorem is verified for f(x) on (1, 4)

Question 5.
Verify Mean Value Theorem, if f (x) = x³ – 5x² – 3x in the interval [a, b], where a = 1 and b = 3. Find all c ∈ (1, 3) for which f’ (c) = 0.
Solution:
f (x) = x³ – 5x² – 3x
f'(x) = 3x² – 10x – 3
Since f'(x’) exists, f(x) is continous on [1, 3]
f(x) is differentiable on (1, 3)
f'(c) = 3c² – 10c – 3
f(b) = f(3) = – 27
f(a) = f(1) = – 7
∴ f'(c) = 0 \(\frac{f(b)-f(a)}{b-a}\)
⇒ 3c² – 10c – 3 = \(\frac{-27-7}{3-1}\)
⇒ 3c² – 10c – 3 = – 10
⇒ 3c² – 10c + 7 = 0
⇒ (c – 1)(3c – 7) = 0 ⇒ c = 1 or c = \(\frac{7}{3}\)
\(\frac{7}{3}\) ∈ (1, 3)
∴ Mean Value Theorem is verified for f(x) on (1, 3)

Question 6.
Examine the applicability of Mean Value theroem for all three functions given in the above Question 2.
Solution:
(i) F (x)= [x] for x ∈ [5, 9], f (x) = [x] in the interval [5, 9] is neither continuous, nor differentiable.
(ii) f (x) = [x], for x ∈ [-2, 2],
Again f (x) = [x] in the interval [-2, 2] is neither continous, nor differentiable.
(iii) f(x) = x² – 1 for x ∈ [1,2], It is a polynomial.
Therefore it is continuous in the interval [1,2] and differentiable in the interval (1,2)
f (x) = 2x, f(1) = 1 – 1 = 0 ,
f(2) = 4 – 1 = 3, f'(c) = 2c
∴ f'(c) = 0 \(\frac{f(b)-f(a)}{b-a}\)
2c = \(\frac{3-0}{2-1}\) = \(\frac{3}{1}\)
∴ c = \(\frac{3}{2}\) which belong to (1, 2)

These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-5-ex-5-7/

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.7

Ex 5.7 Class 12 Maths Ncert Solutions Question 1.
x² + 3x + 2
Solution:
Let y = x² + 3x + 2
Differentiating both sides w.r.t. x
\(\frac { dy }{ dx }\) = 2x + 3 and \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) = 2

Ex 5.7 Class 12 NCERT Solutions Question 2.
x²⁰ = y
Solution:
Let y = x²⁰
Differentiating both sides w.r.t. x
\(\frac { dy }{ dx } ={ 20 }x^{ 19 }\quad ⇒\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =20\times { 19x }^{ 18 }={ 380 }x^{ 18 }\qquad \)

Question 3.
x.cos x = y(say)
Solution:
Let y = x.cos x
Differentiating both sides w.r.t. x
\(\frac { dy }{ dx } \) = x(- sinx) + cosx.1 = – xsinx + cosx
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) = – xcosx – sinx – sinx = – xcosx – 2sinx

Question 4.
log x = y (say)
Solution:
Let y = log x
Differentiating both sides w.r.t. x
\(\frac { dy }{ dx } =\frac { 1 }{ x } ⇒ \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-\frac { 1 }{ { x }^{ 2 } } \)

Question 5.
x³ log x = y (say)
Solution:
Let y = x³ log x
Differentiating both sides w.r.t. x
\( ⇒\frac { dy }{ dx } ={ x }^{ 3 }.\frac { 1 }{ x } +logx\times { 3x }^{ 2 }={ x }^{ 2 }+{ 3x }^{ 2 }logx \)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =2x+{ 3x }^{ 2 }.\frac { 1 }{ x } +logx.6x=x(5+6logx) \)

Question 6.
e^x sin 5x = y
Solution:

Question 7.
e^6x cos3x
Solution:
Let y = e^6x cos3x
Differentiating both sides w.r.t. x

Question 8.
tan^-1 x
Solution:
Let y = tan^-1 x
Differentiating both sides w.r.t. x
\(\frac { dy }{ dx } =\frac { 1 }{ 1+{ x }^{ 2 } } ⇒\frac { { d }^{ 2y } }{ { dx }^{ 2 } } =\frac { -2x }{ { ({ 1+x }^{ 2 }) }^{ 2 } } \)

Question 9.
log(logx)
Solution:
Let y = log(logx)
Differentiating both sides w.r.t. x

Question 10.
sin(log x)
Solution:
Let y = sin(log x)
Differentiating both sides w.r.t. x

Question 11.
If y = 5 cos x – 3 sin x, prove that \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +y=0\)
Solution:
Let y = 5 cos x – 3 sin x
Differentiating both sides w.r.t. x
\(\frac { dy }{ dx } \) = – 5sinx – 3cosx
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) = – 5cosx +3sinx = – y
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) + y = 0
Hence proved

Question 12.
If y = cos^-1 x, Find \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) in terms of y alone.
Solution:
Let y = cos^-1 x
Differentiating both sides w.r.t. x
\(\frac { dy }{ dx } = – { \left( { 1-x }^{ 2 } \right) }^{ -\frac { 1 }{ 2 } }\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\frac { -cosy }{ { \left( { sin }^{ 2 }y \right) }^{ \frac { 3 }{ 2 } } } = -coty\quad { cosec }^{ 2 }y\)

Question 13.
If y = 3 cos (log x) + 4 sin (log x), show that
\({ x }^{ 2 }{ y }_{ 2 }+{ xy }_{ 1 } \) + y = 0
Solution:
Let y = 3 cos (log x) + 4 sin (log x)
Differentiating both sides w.r.t. x

Question 14.
If A = Ae^mx + Be^nx, show that \(\frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y\) = 0
Solution:
Let y = Ae^mx + Be^nx
Differentiating both sides w.r.t. x
\(\frac { d }{ dx }\) = Ae^mx + Be^nx
Differentiating both sides w.r.t. x
\(\frac{d^{2} y}{d x^{2}}\) = Ae^mx(m) + Be^nx(n)
\(\frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y\)
= Ae^mx . (m²) + Be^nx(n²) – (m + n) [Ae^mx(m) + Be^nx(n)] + mny
= Am²e^mx + Bn²e^nx – [Am²e^mx – Bn²e^nx +mny
= – mn[Ae^mx + Be^nx] + mny
= – mny + mny = 0 = RHS

Question 15.
If y = 500e^7x + 600e^-7x, show that \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) = 49y.
Solution:
Let y = 500e^7x + 600e^-7x
Differentiating both sides w.r.t. x

Question 16.
If e^y(x+1) = 1, show that \(\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}\)
Solution:
Let y = e^y(x+1)
Differentiating both sides w.r.t. x

Question 17.
If y = (tan^-1 x)² show that (x² + 1)²y₂ + 2x(x² + 1)y₁ = 2
Solution:
Let y = (tan^-1 x)²
Differentiating both sides w.r.t. x

These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-5-miscellaneous-exercise/

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Miscellaneous Exercise

Question 1.
(3x² -9x + 5)⁹
Solution:
Let y = (3x² -9x + 5)⁹
∴ \(\frac { dy }{ dx }\) = (3x² -9x + 5)⁸. \(\frac { d }{ dx }\)(3x² -9x + 5)
= 9(3x² -9x + 5)⁸ (6x – 9)
= 27 (3x² -9x + 5)⁸ (2x – 3)

Question 2.
sin³x + cos⁶ x
SoL
Let y = sin³x + cos⁶ x
∴ \(\frac { dy }{ dx }\) = 3 sin²x . cosx + 6cos5x (- sinx)
= 3 sinx cosx (sinx – 2 cos⁴x)

Question 3.
(5x)^3cos2x
Solution:
Let y = (5x)^3cos2x
Taking logarithmon both sides,
∴ log y = 3 cos 2x log 5x
Differentiating both sides, w.r.t. x,

Question 4.
sin^-1 (x\(\sqrt{x}\)), 0 ≤ x ≤ 1
Solution:

Question 5.
\(\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}},-2<x<2\)
Solution:

Question 6.
\(\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right], 0<x<\frac{\pi}{2}\)
Solution:

Question 7. \((\log x)^{\log x}, x>1\)
Solution:
Let y = \((\log x)^{\log x}\)
Taking logarithmon both sides,
∴ log y = log x(log log x)
Differentiating both sides, w.r.t. x,
\(\frac { 1 }{ y }\) \(\frac { dy }{ dx }\) = logx.\(\frac { 1 }{ log x }\).\(\frac { 1 }{ x }\) + \(\frac { 1 }{ x }\).log log x
∴ \(\frac { dy }{ dx }\) = (log x)^{log x} [\(\frac { 1 }{ x }\) + \(\frac { log log.x }{ x }\)]

Question 8.
cos (a cos x + b sin x), for sorne constant a and b.
Solution:
Let y = cos (a cosx + b sinx)
\(\frac { dy }{ dx }\) = sin (a cosx + h sinx).
\(\frac { dy }{ dx }\)(a cosx + b sinx)
= – sin (a cosx + b sinx) [- a sinx + b cosx]
= (a sinx – b cosx).sin (a cosx + b sinx)

Question 9.
(sin x – cos x)^{sin x-cos x}, \(\frac { π }{ 4 }\)< x < \(\frac { 3π }{ 4 }\)
Solution:
When \(\frac { π }{ 4 }\)< x < \(\frac { 3π }{ 4 }\), then sin x > cos x
so that sin x – cos x is positive.
Let y = (sinx – cosx)^{sin x-cos x}
Taking logarithm on both sides,
∴ logy = (sinx – cosx) log (sinx – cosx)
Differentiating both sides w.r.t. x,
\(\frac { 1 }{ y }\) \(\frac { dy }{ dx }\)
=( sinx – cosx)\(\frac { 1 }{ sin x-cosx }\) .\(\frac { d }{ dx }\)(sinx-cosx) + log (sinx – cosx). (cosx + sinx)
= (cosx + sinx) + log (sinx – cosx) . (cosx + sinx)
= (cosx + sinx) [1 + log (sinx – cosx)]
∴ \(\frac { dy }{ dx }\) = (sin x – cos x)^{sin x-cos x}(cosx + sinx) [1 + log(sinx – cosx)]

Question 10.
x^x + x^a + a^x + a^a for some fixed a > 0 and x > 0.
Solution:

Question 11.
\(x^{x^{2}-3}+(x-3)^{x^{2}}\), for x > 3
Solution:
Let u = x\(x^{x^{2}-3}\) and v = (x – 3)^x²
Differentiating w.r.t. x,
∴ \(\frac { dy }{ dx }\) = \(\frac { du }{ dx }\) + \(\frac { dv }{ dx }\) … (1)
u = x^x²-3
Taking logarithm on bath sides,
∴ log u = (x² – 3)log
Differentiating both sides w.r.t. x,

Taking logarithm on bath sides,
∴ log v = x²log(x – 3)
Differentiating both sides w.r.t. x,

Question 12.
Find \(\frac { dy }{ dx }\), if y = 12(1 – cos t),
x = 10(t – sin t), \(\frac { – π }{ 2 }\) < t < \(\frac { π }{ 2 }\)
Solution:

Question 13.
Find \(\frac { dy }{ dx }\), if
y = sin^-1 x + sin^-1 \(\sqrt{1-x^{2}},-1 \leq x \leq 1\).
Solution:

Question 14.
If \(x \sqrt{1+y}+y \sqrt{1+x}\) = 0, for – 1 < x < 1, prove that \(\frac { dy }{ dx }\) = – \(\frac{1}{(1+x)^{2}}\)
Solution:
\(x \sqrt{1+y}+y \sqrt{1+x}\) = 0
\(x \sqrt{1+y}\) = – y\(\sqrt{1+y}\)
Squaring both sides,
x²(1 + y) = y²(1 + x)
x² + x²y = y² + y²x
x² – y² = y²x – x²y
(x – y)(x + y) = xy(x – y)
(x + y) = – xy
y + xy = – x
y(1 + x) = – x
∴ y = \(\frac { – x }{ 1+x }\)
Differentiating w.r.t. x,
\(\frac { dy }{ dx }\) = \(\frac{(1+x)(-1)-(-x) 1}{(1+x)^{2}}\)
= \(\frac{-1-x+x}{(1+x)^{2}}\) = \(\frac{-1}{(1+x)^{2}}\), x ≠ – 1

Question 15.
If (x – a)² + (y – b)² = c², for some c > 0, prove that
\(\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}}\)
is a constant, independent of a and b.
Solution:

Question 16.
If cos y = x cos (a + y), with
cos a ≠ ± 1, prove that \(\frac { dy }{ dx }\) = \(\frac{\cos ^{2}(a+y)}{\sin a}\)
Solution:

Question 17.
If x = a(cos t + t sin t) and y = a(sin t – t cos t), find \(\frac{d^{2} y}{d x^{2}}\)
Solution:

Question 18.
If f(x) = |x|³, show that f”(x) exists for all real x and find it.
Solution:
f(x) can be redefined as
f(x) = \(\left\{\begin{aligned}
x^{3}, & x \geq 0 \\
-x^{3}, & x<0 \end{aligned}\right.\) For x > 0 and x < 0, fix) is a polynomial function. Hence f(x) is differentiable for x > 0 and x < 0. ∴ For x > 0, f'(x) = 3x² and f”(x) = 6x
For x < 0, f'(x) = – 3x² and f”(x) = – 6x

or f”(x) = 6|x| exists for all x ∈ R

Question 19.
Using mathematical induction, prove that \(\frac { d }{ dx }\)(xⁿ) = \(n x^{n^{-1}}\) for all positive integers n.
Solution:

Hence P(k + 1) is true.
i.e., P(k + 1) is true whenever P(k) is true.
Hence by the principle of mathematical induction, \(\frac { d }{ dx }\)(xⁿ) = nx^n-1 is true for positive integer n.

Question 20.
Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation, obtain the sum formula for cosines.
Solution:
sin (A + B) = sin A cos B + cos A sin B
Differentiating both sides w.r.t. x,

Question 21.
Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.
Solution:
Yes.
i. Let f(x) = |x – 1| + |x – 2|
Let g(x) = |x|, h(x) = x – 1, k(x) = x – 2.
Then g(x), h(x) and k(x) are continuous functions since h(x) and k(x) are polynomial functions and g(x) ¡s the modulus function.

at x = 1, since Lf’(1) ≠ Rf’(1)
Similarly we can show that f(x) ¡s not differentiable at x = 2.
Thus f(x) = |x – 1| + |x – 2| is continuous everywhere and not differentiable at exactly two points, namely at x = 1 or x = 2.

Question 22.
If y = \(\left|\begin{array}{ccc}
f(x) & g(x) & h(x) \\
l & m & n \\
a & b & c
\end{array}\right|\), prove that \(\frac { dy }{ dx }\) = \(\left|\begin{array}{ccc}
f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\
l & m & n \\
a & b & c
\end{array}\right|\)
Solution:

Question 23.
If y = \(e^{a \cos ^{-1} x}\), – 1 ≤ x ≤ 1, show that (1 – x²)\(\frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-a^{2} y\) = 0
Solution:
f(x) = x + \(\frac { 1 }{ x }\)
f(x) is a continuous function in [1, 3]
Ax) is differentiable in(1,3)
f’(x) = 1 + \(\\frac{-1}{x^{2}}\) exists for x ∈ (1, 3)

These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-5-ex-5-4/

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.4

Ex 5.4 Class 12 NCERT Solutions Question 1.
\(\frac { { e }^{ x } }{ sinx } \)
Solution:
\(y=\frac { { e }^{ x } }{ sinx } \)
\(for\quad y=\frac { u }{ v } ,\)
\(\frac { dy }{ dx } =\frac { { e }^{ x }{ sin }x-{ e }^{ x }cosx }{ { sin }^{ 2 }x } \)
\(or\frac { dy }{ dx } =\frac { { e }^{ x }{ sin }x-{ e }^{ x }cosx }{ { sin }^{ 2 }x } ,where\quad x\neq n\pi ,x\in z \)

Exercise 5.4 Class 12 Maths NCERT Solutions Question 2.
\({ e }^{ { sin }^{ -1 }x }\)
Solution:

Question 3.
\({ e }^{ { x }^{ 3 } }\) = y
Solution:
Let y = \({ e }^{ { x }^{ 3 } }\)
Differentiating w.r.t. x,
\(\frac{d y}{d x}=\frac{d}{d x}\left(e^{x^{3}}\right)\) = \(e^{x^{3}} \cdot \frac{d}{d x}\left(x^{3}\right)\)
= \({ e }^{ { x }^{ 3 } }\).3x² = 3x²\({ e }^{ { x }^{ 3 } }\)

Question 4.
\(sin\left( { tan }^{ -1 }{ e }^{ -x } \right)\) = y
Solution:
\(sin\left( { tan }^{ -1 }{ e }^{ -x } \right)\) = y
\(\frac { dy }{ dx } =cos\left( { tan }^{ -1 }{ e }^{ -x } \right) \frac { d }{ dx } \left( { tan }^{ -1 }{ e }^{ -x } \right) \)
\(=cos\left( { tan }^{ -1 }{ e }^{ -x } \right) \frac { 1 }{ 1+{ e }^{ -2x } } \frac { d }{ dx } \left( { e }^{ -x } \right) \)
\(=-cos\left( { tan }^{ -1 }{ e }^{ -x } \right) \frac { 1 }{ 1+{ e }^{ -2x } } .\left( { e }^{ -x } \right) \)

Question 5.
\(log(cos\quad { e }^{ x })\) = y
Solution:
\(\frac { dy }{ dx } =\frac { 1 }{ cos\quad { e }^{ x } } \left( -sin{ e }^{ x } \right) .{ e }^{ x }\quad =-tan\left( { e }^{ x } \right) \)

Question 6.
\({ e }^{ x }+{ e }^{ { x }^{ 2 } }+\)…\(+{ e }^{ { x }^{ 5 } }\) = y(say)
Solution:

Question 7.
\(\sqrt { { e }^{ \sqrt { x } } }\), x > 0
Solution:

Question 8.
log(log x), x > 1
Solution:
y = log(log x), x > 1
Differentiating w.r.t. x,
\(\frac{d y}{d x}\) = \(\frac{1}{\log x} \cdot \frac{d}{d x}\)(log x)
= \(\frac{1}{\log x} \cdot \frac{1}{x}\) = \(\frac{1}{x \log x}\), x > 1

Question 9.
\(\frac { cosx }{ logx }\) = y(say),x>0
Solution:

Question 10.
cos(log x + ex), x > 0
Solution: