# MP Board Class 12th Maths Important Questions Chapter 5B Differentiation

## MP Board Class 12th Maths Important Questions Chapter 5B Differentiation

### Differentiation Important Questions

Question 1.
Differentiate the function sin(cos x2) with respect to x? (NCERT)
Solution:
Let y = sin (cosx2)
$$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$ sin (cos x2)
= $$\frac { d }{ dx }$$ sin t, [Putting cos x2 = t]
= $$\frac { d }{ dt }$$ sin t $$\frac { dt }{ dx }$$
= cos t $$\frac { d }{ dx }$$ cos x2
= cos (cos x2) $$\frac { d }{ dx }$$ cos u, [Putting x2 = u]
= cos (cos x2) $$\frac { d }{ du }$$ cos u $$\frac { du }{ dx }$$
= – cos (cos x2) sin u $$\frac { d }{ dx }$$ x2
= – 2x cos (cos x2). sin x2.

Question 2.
Differentiate the function y = sec [tan $$\sqrt { x }$$ ] with respect to x? (NCERT)
Solution:
Given:
y = sec [tan $$\sqrt { x }$$ ]
$$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$ sec [tan $$\sqrt { x }$$ ]
= $$\frac { d }{ dx }$$ sec t, [Putting tan $$\sqrt { x }$$ = t]
= $$\frac { d }{ dt }$$ sec t $$\frac { dt}{ dx }$$
= sec t tan t $$\frac { d }{ dx }$$ tan $$\sqrt { x }$$
= sec (tan $$\sqrt { x }$$) tan (tan $$\sqrt { x }$$) $$\frac { d }{ dx }$$ tan u, [Putting $$\sqrt { x }$$ = u]
= sec (tan $$\sqrt { x }$$) tan (tan $$\sqrt { x }$$) sec2 u $$\frac { d }{ dx }$$ $$\sqrt { x }$$
= sec (tan $$\sqrt { x }$$) tan (tan $$\sqrt { x }$$) sec2$$\sqrt { x }$$ × $$\frac { 1 }{ 2\sqrt { x } }$$

Question 3.
Differentiate the function y = log [cos ex] with respect to x? (NCERT)
Solution:
Given:
y = log [cos ex]
$$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$ [log (cos ex)]
$$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$ log t, [Putting cos ex = t]
= $$\frac { d }{ dt }$$ log t $$\frac { dt }{ dx }$$
= $$\frac { 1 }{ t }$$. $$\frac { d }{ dx }$$ cos ex
= $$\frac { 1 }{ cose^{ x } }$$ × $$\frac { d }{ dx }$$ cos u, [Putting ex = u]
= $$\frac { 1 }{ cose^{ x } }$$. $$\frac { d }{ du }$$ cos u $$\frac { du }{ dx }$$
= $$\frac { -sinu }{ cose^{ x } }$$. $$\frac { d }{ dx }$$ ex
= $$\frac { -(sine^{ x })e^{ x } }{ cose^{ x } }$$
= – ex tan ex

Question 4.
Differentiate the function y = cos [log x + ex] with respect to x? (NCERT)
Given:
y = cos [log x + ex]
$$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$ cos (log x + ex)
$$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$ cos t, [Putting log x + ex = t]
= $$\frac { d }{ dt }$$ cos t $$\frac { dt }{ dx }$$
= – sin t $$\frac { d }{ dx }$$ (log x + ex)
= – sin (log x + ex) ($$\frac{1}{x}$$ + ex)
= – $$\frac { (xe^{ x }+1)sin(logx+e^{ x }) }{ x }$$

Question 5.
Differentiate the function y = cos-1(ex) with respect to x? (NCERT)
Solution:
Given:
y = cos-1 (ex)
∴$$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$ cos-1 (ex)
Putting ex = t,
= $$\frac { d }{ dx }$$ cos-1 t = $$\frac { d }{ dt }$$ cos-1 t $$\frac { dt }{ dx }$$
= $$\frac { 1 }{ \sqrt { 1-t^{ 2 } } }$$ $$\frac { d }{ dx }$$ ex
= – $$\frac { e^{ x } }{ \sqrt { 1-e^{ 2x } } }$$

Question 6.
If y + sin y = cos x then find the value of $$\frac { dy }{ dx }$$? (NCERT)
Solution:
Given:
y + sin y = cos x
Differentiating both sides with respect to x,
$$\frac { d }{ dx }$$ (y + siny) = $$\frac { d }{ dx }$$ cos x
⇒ $$\frac { dy }{ dx }$$ + cos y $$\frac { dy }{ dx }$$ = – sin x
⇒ $$\frac { dy }{ dx }$$ (1 + cos y) = -sin x
⇒ $$\frac { dy }{ dx }$$ = $$\frac { -sinx }{ 1+cosy }$$

Question 7.
If 2x + 3y = sin x then find the value of $$\frac { dy }{ dx }$$? (NCERT)
Solution:
Given:
2x + 3y = sin x
Differentiating both sides with respect to x,
$$\frac { d }{ dx }$$ (2x + 3y) = $$\frac { d }{ dx }$$ sin x
2 $$\frac { d }{ dx }$$ x + 3 $$\frac { dy }{ dx }$$ = cos x
⇒ 2 + 3 $$\frac { dy }{ dx }$$ = cos x – 2
∴ $$\frac { dy }{ dx }$$ = $$\frac{cos x – 2}{3}$$

Question 8.
If x = a cos θ, y = a sin θ then find the value of $$\frac { dy }{ dx }$$? (NCERT)
Solution:
Given:
x = a cos θ
y = a sin θ
Differentiating eqn. (1) with respect to θ.
We get, $$\frac { dx }{ d\theta }$$ = – a sin θ
Again, $$\frac { dy }{ dx }$$ = $$\frac { \frac { dy }{ d\theta } }{ \frac { dx }{ d\theta } }$$
⇒ $$\frac { dy }{ dx }$$ = – $$\frac { acos\theta }{ asin\theta }$$
⇒ $$\frac { dy }{ dx }$$ = – cot θ.

Question 9.
If x = at2 and y = 2at then find the value of $$\frac { dy }{ dx }$$? (NCERT)
Solution:
Given:
x = at2
$$\frac { dx }{ dt }$$ = 2at
y = 2at
$$\frac { dy }{ dt }$$ = 2a
Again, $$\frac { dy }{ dx }$$ = $$\frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } }$$ = $$\frac{2a}{2at}$$
⇒ $$\frac { dy }{ dx }$$ = $$\frac { 1 }{ t}$$.

Question 10.
If y = x2 + 3x + 2 then find the value of $$\frac { d^{ 2 }y }{ dx^{ 2 } }$$? (NCERT)
Solution:
Given:
y = x2 + 3x + 2
∴ $$\frac { dy }{ dx }$$ = 2x + 3.1 + 0
$$\frac { dy }{ dx }$$ = 2x + 3
Again differentiating with respect to x,
We get, $$\frac { d }{ dx }$$ ( $$\frac { dy }{ dx }$$ ) = 2.1 + 0
⇒ $$\frac { d^{ 2 }y }{ dx^{ 2 } }$$ = 2.

Question 11.
If y = x3 + tan x then find the value of $$\frac { d^{ 2 }y }{ dx^{ 2 } }$$? (NCERT)
Solution:
Given:
y = x3 + tan x
$$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$ [x3 + tan x]
= $$\frac { d }{ dx }$$ x3 + $$\frac { d }{ dx }$$ tan x
⇒ $$\frac { dy }{ dx }$$ = 3x2 + sec2 x
Again, differentiating with respect to x,
⇒ $$\frac { d }{ dx }$$ ( $$\frac { dy }{ dx }$$ ) = $$\frac { d }{ dx }$$ [3x2 + sec2x]
$$\frac { d^{ 2 }y }{ dx^{ 2 } }$$ = 3 $$\frac{d}{dx}$$ x2 + $$\frac{d}{dx}$$ sec2 x
⇒ $$\frac { d^{ 2 }y }{ dx^{ 2 } }$$ = 6x + $$\frac{d}{dx}$$ t2, [Putting sec x = t]
= 6x + $$\frac { d }{ dt }$$ t2 $$\frac { dt }{ dx }$$
= 6x + 2t $$\frac { d }{ dx }$$ sec x
= 6x + 2 sec x.secx tanx
⇒ $$\frac { d^{ 2 }y }{ dx^{ 2 } }$$ = 6x + 2 sec2x tan x.

Differentiation Long Answer Type Questions – I

Question 1.
If y = tan-1 $$\frac { x }{ \sqrt { 1+x^{ 2 } } }$$ then find the value of $$\frac { dy }{ dx }$$?
Solution:
Given:
y = tan-1 $$\frac { x }{ \sqrt { 1+x^{ 2 } } }$$
Now putting $$\frac { x }{ \sqrt { 1+x^{ 2 } } }$$ = t
$$\frac { dy }{ dx }$$ = $$\frac { d }{ dt }$$ tan-1t. $$\frac { dt }{ dx }$$
= $$\frac { 1 }{ 1+t^{ 2 } }$$. $$\frac { d }{ dx }$$ $$\frac { x }{ \sqrt { 1+x^{ 2 } } }$$

Again Putting 1 + x2 = u,

Question 2.
If x = a (t + sin t) and y = a(1 – cost) then find the value of $$\frac { dy }{ dx }$$?
Solution:
Given:
x = a (t + sin t)
∴$$\frac { dx }{ dt }$$ = a(1 + cos t)
Again y = a (1 – cos t)
∴$$\frac { dy }{ dt }$$ = a (0 + sint) = a sin t
Hence $$\frac { dy }{ dx }$$ = $$\frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } }$$ = $$\frac { asint }{ a(1+cost) }$$
= $$\frac { sint }{ a(1+cost) }$$ = $$\frac { 2sint/2cost/2 }{ 2cos^{ 2 }t/2 }$$
⇒ $$\frac { dy }{ dx }$$ = tan $$\frac{t}{2}$$.

Question 3.
If x = a(2θ – sin 2θ) and y = a(1 – cos 2θ) then find the value of $$\frac { dy }{ dx }$$ where θ = $$\frac { \pi }{ 3 }$$? (CBSE 2018)
Solution:
Given:
x = a (2θ – sin 2θ) ………………… (1)
y = a (1 – cos 2θ) ………………………. (2)
Differentiating eqn. (1) with respect to θ, we get
$$\frac { dx }{ d\theta }$$ = a(2.1 – cos 2θ.2)
= 2a (1 – cos 2θ)
= 2a.2 sin2θ
= 4a sin2θ
Differentiating eqn. (2) with respect to θ, ………………… (3)
$$\frac { dy }{ d\theta }$$ = a (0 + sin 2θ.2)
= 2a sin 2θ
= 2a.2 sin θ cos θ ……………………….. (4)
= 4a sin θ cos θ
Divinding eqn.(4) by eqn.(3),
$$\frac { dy }{ d\theta }$$ + $$\frac { dx }{ d\theta }$$ = $$\frac { 4asin\theta cos\theta }{ 4asin^{ 2 }\theta }$$
⇒ $$\frac { dy }{ dx }$$ = cot θ
When θ = $$\frac { \pi }{ 3 }$$, then
$$\frac { dy }{ dx }$$ = cot $$\frac { \pi }{ 3 }$$ = $$\frac { 1 }{ \sqrt { 3 } }$$.

Question 4.
If y = a sin mx + b cos mx then prove that:
$$\frac { d^{ 2 }y }{ dx^{ 2 } }$$ + m2y = 0?
Solution:
Given:
y = a sin mx + b cos mx ……………………. (1)
Differentiating eqn. (2) with respect to x,
$$\frac { dy }{ dx }$$ = am cos mx – bm sin mx
Differentiating eqn. (2) with respect to x,
$$\frac { d^{ 2 }y }{ dx^{ 2 } }$$ = – am2 sin mx – bm2 cos mx
⇒ $$\frac { d^{ 2 }y }{ dx^{ 2 } }$$ = – m2 y
∴ $$\frac { d^{ 2 }y }{ dx^{ 2 } }$$ + m2 y = 0. Proved.

Question 5.
(A) If y = emsin-1x then prove that (1 – x2) y2 – xy1 – m2y = 0?
Solution:
Given:
y = emsin-1x
$$\frac { dy }{ dx }$$ = y1 = emsin-1x. $$\frac { d }{ dx }$$ (msin-1x)
⇒ y1 = y.m. $$\frac { 1 }{ \sqrt { 1-x^{ 2 } } }$$
⇒ $$\sqrt { 1-x^{ 2 } }$$ y1 = my …………………………. (1)
Again, differentiating with respect to x,
$$\sqrt { 1-x^{ 2 } }$$. y2 + y1. $$\frac{1}{2}$$ (1 – x2)1/2 (- 2x) = my1
⇒ $$\sqrt { 1-x^{ 2 } }$$. y2 – $$\frac { x }{ \sqrt { 1-x^{ 2 } } }$$ y1 = m$$\frac { my }{ \sqrt { 1-x^{ 2 } } }$$. [from eqn.(1)]
⇒ (1 – x2) y2 – xy1 = m2y
⇒ (1 – x2) y2 – xy1 – m2y = 0. Proved.

Question 5.
(B) If y = emtan-1x then prove that (1 + x2) y2 + (2x – m) y1 = 0?
Solution:
Solve like Q.5 (A)

Question 5.
(C) If y = emcos-1x then prove that (1 – x2) y2 – xy1 – m2 y = 0?
Solution:
Solve like Q.5 (A)

Question 6.
Differentiate sin-1 $$\frac { 2x }{ 1+x^{ 2 } }$$ with respect to x?
Solution:
Let y = sin-1 ( $$\frac { 2x }{ 1+x^{ 2 } }$$ )
Again let x = tan θ ⇒ θ = tan-1 x
y = sin-1 ( $$\frac { 2tan\theta }{ 1+tan^{ 2 }\theta }$$ )
= sin-1 (sin 2θ) = 2θ = 2 tan-1 x
∴$$\frac { dy }{ dx }$$ = 2 $$\frac { d }{ dx }$$ (tan-1 x) = $$\frac { 2 }{ 1+x^{ 2 } }$$

Question 7.
If y = cot-1 $$\sqrt { \frac { 1+x }{ 1-x } }$$ then find $$\frac { dy }{ dx }$$?
Solution:
Given:
y = cot-1 $$\sqrt { \frac { 1+x }{ 1-x } }$$
Let x = cos θ

Putting in eqn.(1), we get
y = cot-1 (cot $$\frac{θ}{2}$$)
⇒ y = $$\frac{θ}{2}$$ = $$\frac{1}{2}$$ cos-1 x, [∵x = cos θ ⇒ ∴θ = cos-1 x]
Differentiating both sides w.r.t. x,
$$\frac { dy }{ dx }$$ = – $$\frac { 1 }{ 2\sqrt { 1-x^{ 2 } } }$$

Question 8.
If y = tan-1 $$\sqrt { \frac { 1+x }{ 1-x } }$$ then find $$\frac { dy }{ dx }$$?
Solution:
Solve like Q.No.7
$$\frac { dy }{ dx }$$ = $$\frac { 1 }{ 2\sqrt { 1-x^{ 2 } } }$$

Question 9.
If y = cot-1 ( $$\frac { cosx+sinx }{ cosx-sinx }$$ ) then find the value of $$\frac { dy }{ dx }$$?
Solution:
Given:
y = cot-1 ( $$\frac { cosx+sinx }{ cosx-sinx }$$ )

⇒ $$\frac { dy }{ dx }$$ = $$\frac { d }{ dx }$$ ( $$\frac { \pi }{ 4 }$$ – x) = -1.

Question 10.
y = tan-1 $$\frac { \sqrt { 1+x^{ 2 }-1\quad } }{ x }$$ Differentiate with respect to x?
Solution:
Given:
y = tan-1$$\frac { \sqrt { 1+x^{ 2 }-1\quad } }{ x }$$ ……………….. (1)
Put x = tan θ in eqn. (1)
∴ θ = tan-1 x

⇒ y = $$\frac { \theta }{ 2 }$$ = $$\frac{1}{2}$$ tan-1 x
∴$$\frac { dy }{ dx }$$ = $$\frac{1}{2}$$ tan-1 x
∴$$\frac { dy }{ dx }$$ = $$\frac{1}{2}$$ $$\frac { d }{ dx }$$ (tan-1 x ) = $$\frac{1}{2}$$ $$\frac { 1 }{ (1+x^{ 2 }) }$$

Question 11.
If y = cot-1 $$\left[\frac{\sqrt{1+x^{2}}+1}{x}\right]$$ then find the value of $$\frac { dy }{ dx }$$?
Solution:
y = cot-1 $$\left[\frac{\sqrt{1+x^{2}}+1}{x}\right]$$
Put x = tan θ,

⇒ y = $$\frac{1}{2}$$ tan-1x
⇒ $$\frac { dy }{ dx }$$ = $$\frac{1}{2}$$. $$\frac { 1 }{ 1+x^{ 2 } }$$.

Question 12.
If y = xsinx then find the value of $$\frac { dy }{ dx }$$?
Solution:
Given:
y = xsinx
Taking log on both sides with respect to x.
$$\frac { 1 }{ y }$$ $$\frac { dy }{ dx }$$ = sin x × $$\frac{1}{x}$$ + logx cos x
∴$$\frac { dy }{ dx }$$ = y.[ $$\frac{sinx}{x}$$ + log x.cos x]

Question 13.
If y = $$\sqrt { \frac { 1-x }{ 1+x } }$$ then prove that $$\frac { dy }{ dx }$$ = $$\frac { y }{ x^{ 2 }-1 }$$?
Solution:
Given:
y = $$\sqrt { \frac { 1-x }{ 1+x } }$$1/2
By taking log , log y = log $$\sqrt { \frac { 1-x }{ 1+x } }$$1/2
⇒ log y = $$\frac{1}{2}$$ [log (1 – x) – log (1 + x)]
Differentiating both sides with respect to x,

Question 14.
If y = (sin x)sinxsinx ………. ∞ then find the value of $$\frac { dy }{ dx }$$?
Solution:
Given:
y = (sin x)sinxsinx ………. ∞
⇒ y = (sin x)y
⇒ log y = y log sin x
Differentiating both sides with respect to x,
$$\frac{1}{y}$$ $$\frac{dy}{dx}$$ = y $$\frac{d}{dx}$$ (log sin x) + log sin x $$\frac{dy}{dx}$$

Question 15.
(A) If y = $$\sin x+\sqrt{\sin x+\sqrt{\sin x+\ldots+\infty}}$$ then prove that:
$$\frac{dy}{dx}$$ = $$\frac{cos x}{2y – 1}$$
Solution:

Differentiating both sides with respect to x,
2y $$\frac{dy}{dx}$$ = cos x + $$\frac{dy}{dx}$$
⇒ 2y $$\frac{dy}{dx}$$ – $$\frac{dy}{dx}$$ = cos x
⇒ (2y – 1) $$\frac{dy}{dx}$$ = cos x
∴$$\frac{dy}{dx}$$ = $$\frac{cos x}{2y – 1}$$.

(B) If y = $$\cot x+\sqrt{\cot x+\sqrt{\cot x+\ldots+\infty}}$$ then prove that:
$$\frac{dy}{dx}$$ = $$\frac { cosec^{ 2 }x }{ 1-2y }$$
Solution:
Solve like Q.No. 15 (A).

(C) If y = \begin{aligned} &x+\sqrt{\tan x+\sqrt{\tan x+\ldots \infty}}\\ \end{aligned} then find the value of $$\frac{dy}{dx}$$?
Solution:
Solve like Q.No 15 (A)

Question 16.
If y = e$$x+e^{x+e^{x+e}}-$$ then prove that:
$$\frac{dy}{dx}$$ = $$\frac{y}{1-y}$$?
Solution:
Given: y = e$$x+e^{x+e^{x+e}}-x$$
⇒ y = ex+y
Taking log on both sides,
log y = log ex+y
log y = x + y
Differentiating both sides with respect to x,
$$\frac{1}{y}$$ $$\frac{dy}{dx}$$ = 1 + $$\frac{dy}{dx}$$
⇒ $$\frac{dy}{dx}$$ ( $$\frac{1}{y}$$ – 1) = 1
⇒ $$\frac{dy}{dx}$$ ( $$\frac{1-y}{y}$$ ) = 1
⇒ $$\frac{dy}{dx}$$ = $$\frac{y}{1-y}$$ Proved.

Question 17.
Differentiate $$\frac { 1 }{ (x+a)(x+b)(x+c) }$$ with respect to x?
Solution:
Let y = $$\frac { 1 }{ (x+a)(x+b)(x+c) }$$
Applying log on both sides,
log y = log 1 – log(x + a) – log (x + b) – log(x + c)
Differentiating both sides with respect to x,
$$\frac{1}{y}$$ $$\frac{dy}{dx}$$ = 0 – $$\frac{1}{x + a}$$ – $$\frac{1}{x + b}$$ – $$\frac{1}{x + c}$$
⇒ $$\frac{dy}{dx}$$ = – y [ $$\frac{1}{x + a}$$ + $$\frac{1}{x + b}$$ + $$\frac{1}{x + c}$$ ]
⇒ $$\frac{dy}{dx}$$ = $$\frac { 1 }{ (x+a)(x+b)(x+c) }$$ × { $$\frac { 1 }{ x+a } +\frac { 1 }{ x+b } +\frac { 1 }{ x+c }$$ }

Question 18.
Differentiate log ( $$\sqrt{x}$$ + $$\frac { 1 }{ \sqrt { x } }$$ ) with respect to x?
Solution:
Let y = log ( $$\sqrt{x}$$ + $$\frac { 1 }{ \sqrt { x } }$$ ) ⇒ y = log ( $$\frac { x+1 }{ \sqrt { x } }$$ )
⇒ y = log (x + 1) – log $$\sqrt{x}$$
⇒ y = log (x + 1) – $$\frac{1}{2}$$ log x
Differentiating both sides with respect to x,
$$\frac{dy}{dx}$$ = $$\frac{d}{dx}$$ log (x + 1) – $$\frac{1}{2}$$. $$\frac{d}{dx}$$ log x
⇒ $$\frac{dy}{dx}$$ = $$\frac{1}{x + 1}$$ – $$\frac{1}{2}$$.$$\frac{1}{x}$$ = $$\frac{2x-x-1}{2x(x+1)}$$
⇒ $$\frac{dy}{dx}$$ = $$\frac{x – 1}{2x(x + 1)}$$.

Question 19.
Differentiate y = tan-1 ( $$\frac { sinx }{ 1+cosx }$$ ) with respect to x?
Solution:
y = tan-1 ( $$\frac { sinx }{ 1+cosx }$$ )

⇒ y = tan-1 (tan $$\frac{x}{2}$$ ) = $$\frac{x}{2}$$
Differentiating both sides with respect to x,
$$\frac{dy}{dx}$$ = $$\frac{d}{dx}$$ ( $$\frac{x}{2}$$ ) = $$\frac{1}{2}$$.

Question 20.
Verify Rolle’s theroem for the function f(x) = x2 interval [-1, 1]. (NCERT)
Solution:
Given:
f(x) = x2, a = – 1, b = 1.

1. f(x) = x2 is a polynomial, hence, f(x) is continous in [-1, 1].
2. f'(x) = 2x exist for every value of x, Hence it is differentiable in (-1, 1).
3. f(-1) = (-1)2 = 1, f(1) = (1)2 = 1.

∴ f(-1) = f(1)
There exists a value c in (-1, 1) such that:
∴ f'(c) = 0
⇒ 2c = 0, [∵f'(x) = 2x]
⇒ c = 0 ∈ (-1, 1)
Hence, Rolle’s theorem is verified. Proved.

Question 21.
Verify Rolle’s theroem for the function f(x) = x2 + 2x – 8, x ∈ [-4, 2]? (NCERT)
Solution:
Given:
f(x) = x2 + 2x – 8, a= – 4, b = 2.

1. f(x) = x2 + 2x – 8 is a polynomial hence f(x) is continous in [-4, 2].
2. f'(x) = 2x + 2 exist for every value of x, hence it is differentiable in (-4, 2).
3. f(-4) = (-4)2 + 2 (-4) – 8

= 16 – 8 – 8 = 0
f(2) = (2)2 + 2 × 2 – 8 = 4 + 4 – 8 = 0
∴ f(-4) = f(2).
There exists a value c in (-4, 2),
∴ f'(c) = 0
⇒ 2c + 2 = 0
⇒ c = – 1 ∈ (-4, 2)
Hence, Rolle’s theorem is verified.

Question 22.
Verify Rolle’s theorem for the function f(x) = 2x3 + x2 – 4x – 2?
Solution:
Given:
f(x) = 2x3 + x2 – 4x – 2 …………………… (1)
We know that polynomial functions are continuous for all real values.
∴ f(x) = o
⇒ 2x3 + x2 – 4x – 2 = 0
⇒ x2 (2x + 1) – 2(2x + 1) = 0
⇒ (x2 – 2) (2x + 1) = 0
⇒ x2 = 2, 2x + 1 = 0
⇒ x = ±$$\sqrt { 2 }$$, x = – $$\frac{1}{2}$$
⇒ x = – $$\sqrt { 2 }$$, $$\sqrt { 2 }$$, $$\frac{-1}{2}$$
∴ Interval [-$$\sqrt { 2 }$$, $$\sqrt { 2 }$$ ].
1. f(x) is continuous in [-$$\sqrt { 2 }$$, $$\sqrt { 2 }$$ ]
2. f'(x) = 6x2 + 2x – 4 is differentiable in [-$$\sqrt { 2 }$$, $$\sqrt { 2 }$$].
3. f(-$$\sqrt { 2 }$$) = 2( $$\sqrt { 2 }$$ )3 + (-$$\sqrt { 2 }$$ ) 2 – 4 (- $$\sqrt { 2 }$$ ) – 2 = 0
and f ( $$\sqrt { 2 }$$ ) = 2( $$\sqrt { 2 }$$ )3 + ( $$\sqrt { 2 }$$ ) 2 – 4( $$\sqrt { 2 }$$ ) – 2 = 0
∴ f(- $$\sqrt { 2 }$$ ) = f( $$\sqrt { 2 }$$ )
There exists a value c in (-$$\sqrt { 2 }$$, $$\sqrt { 2 }$$ )
∴ f'(c) = 0
⇒ 6c2 + 2c – 4 = 0, [∵f'(x) = 6x2 + 2x – 4]
∴ c = $$\frac { -2\pm \sqrt { 2^{ 2 }-4\times 6\times (-4) } }{ 2\times 6 }$$
c = $$\frac { -2\pm \sqrt { 4+96 } }{ 12 }$$
⇒ c = $$\frac { -2\pm 10 }{ 12 }$$
⇒ c = $$\frac{-2-10}{12}$$ and c = $$\frac{-2+10}{12}$$
⇒ c = -1, $$\frac{2}{3}$$ ∈ (- $$\sqrt { 2 }$$, $$\sqrt { 2 }$$ )
Hence, Rolle’s theroem is verified.

Question 23.
Verify Lagrange’s mean value theorem for the function f(x) = x + $$\frac{1}{x}$$ on [1, 3].
Solution:
Given:
f(x) = x + $$\frac{1}{x}$$ = $$\frac { x^{ 2 }+1 }{ x }$$, x ∈ [1, 3]

1. f(x), x ≠ 0 hence it is a continous function in [1, 3].
2. f'(x) = 1 – $$\frac { 1 }{ x^{ 2 } }$$ is differentiable in (1, 3).
3. f(1) = 2 and f(3) = $$\frac{10}{3}$$

Hence, f(1) ≠ f(2)
For mean value theorem,
∴ $$\frac { f(b)-f(a) }{ b-a }$$ = f'(c)
⇒ $$\frac { f(3)-f(1) }{ 3-1 }$$ = 1 – $$\frac { 1 }{ c^{ 2 } }$$
⇒ $$\frac { \frac { 10 }{ 3 } -2 }{ 2 }$$ = 1 – $$\frac { 1 }{ c^{ 2 } }$$
⇒ 1 – $$\frac { 1 }{ c^{ 2 } }$$ = $$\frac{2}{3}$$
⇒ $$\frac { 1 }{ c^{ 2 } }$$ = $$\frac{3-2}{3}$$ = $$\frac{1}{3}$$
⇒ c2 = 3
⇒ c = $$\sqrt{3}$$ = 1.732 ∈ (1, 3)
Hence, Langrange’s mean value theorem is verified. Proved.

Question 24.
Verify Lagrange’s mean value theorem for the following function f(x) = log x on [1, e]?
Solution:
f(x) = logx, x ∈ [1, e], x > 0.
1. As f(x) = log x, x > 0 is a continuous function, hence f(x) is continuous in [1, e],

2. f'(x) = $$\frac{1}{x}$$,
∴ f(x) is differentiable in (1, e).

3. f(1) = log 1 = 0, f(e) = log e = 1.
Now by mean value theorem,
∴ $$\frac{f(e) – f(1)}{e-1}$$ = f'(c)
⇒ $$\frac{1-0}{e-1}$$ = $$\frac{1}{e}$$
⇒ c = e – 1 ∈ (1, e)
Hence, Langrange’s mean value theorem is verified. Proved.

Question 25.
With the help of Langrange’s value thoerem for the function y = $$\sqrt{x-2}$$ in the interval [2, 3]. Find the point where the tangent is parallel to be chord joining the points?
solution:
Given:
f(x) = $$\sqrt{x-2}$$, a = 2, b = 3

1. As f(x) = $$\sqrt{x-2}$$, x ∈ [2, 3] is defined.
∴ f(x) is continous function for [2, 3].

2. f'(x) = $$\frac { 1 }{ 2\sqrt { x-2 } }$$ is defined in interval (2, 3).
∴ f(x) is differentiable in [2, 3]

3. f(2) = 0, f(3) = 1
f(2) ≠ f(3)
Now, by Langrange’s mean value theorem,
∴ $$\frac{f(3)-f(2)}{3-2}$$ = f'(c)
⇒ $$\frac{1-0}{1}$$ = $$\frac { 1 }{ 2\sqrt { c-2 } }$$
⇒ $$\frac { 1 }{ 2\sqrt { c-2 } }$$ = 1
⇒ $$\frac { 1 }{ \sqrt { c-2 } }$$ = 2
⇒ $$\sqrt{c-2}$$ = $$\frac{1}{2}$$
⇒ c – 2 = $$\frac{1}{4}$$
⇒ c = $$\frac{1}{4}$$ + 2 = $$\frac{9}{4}$$ = 2.25 ∈ (2,3)
∴ f(c) = $$\sqrt { \frac { 9 }{ 4 } -2 }$$ = $$\frac{1}{2}$$
Required points ( $$\frac{9}{4}$$, $$\frac{1}{2}$$ ).

Differentiation Long Answer Type Questions – II

Question 1.
Differentiate sin-1 [ $$\frac { 2^{ x+1 } }{ 1+4^{ x } }$$ ] with respect to x? (NCERT)
Solution:
y = sin -1 [ $$\frac { 2^{ x+1 } }{ 1+4^{ x } }$$ ]
⇒ y = sin-1 [ $$\frac { 2.2^{ x } }{ 1+2^{ 2x } }$$ ]
Putting 2x = tan θ
Then, θ = tan-1 2x
⇒ y = sin-1 [ $$\frac { 2tan\theta }{ 1+tan^{ 2 }\theta }$$ ]
⇒ y = sin-1 [sin 2θ], [∵sin 2θ = $$\frac { 2tan\theta }{ 1+tan^{ 2 }\theta }$$ ]
⇒ y = 2θ
⇒ y = 2 tan-1 (2x), [θ = tan-1(2x)]
∴ $$\frac{dy}{dx}$$ = 2 $$\frac{d}{dx}$$ tan-1 (2x)
Putting 2x = t
⇒ $$\frac{dy}{dx}$$ = 2 $$\frac{d}{dx}$$ tan-1 t
= 2 $$\frac{d}{dt}$$ tan-1 t$$\frac{dt}{dx}$$
⇒ $$\frac{dy}{dx}$$ = $$\frac { 2 }{ 1+t^{ 2 } }$$ $$\frac{d}{dx}$$ (2x),
= $$\frac { 2 }{ 1+2^{ 2x } }$$ × 2x log 2
⇒ $$\frac{dy}{dx}$$ = $$\frac { 2^{ x+1 }log2 }{ 1+4^{ x } }$$

Question 2.
If y = sin-1 x then prove that: (NCERT)
(1 – x2) $$\frac { d^{ 2 }y }{ dx^{ 2 } }$$ – x $$\frac{dy}{dx}$$ = 0? (NCERT)
Solution:
Given:
y = sin-1 x ……………………………. (1)
$$\frac{dy}{dx}$$ = $$\frac{d}{dx}$$ (sin-1 x)
$$\frac{dy}{dx}$$ = $$\frac { 1 }{ \sqrt { 1-x^{ 2 } } }$$
$$\frac{d}{dx}$$ ( $$\frac{dy}{dx}$$ ) = $$\frac{d}{dx}$$ t-1/2
Putting 1 – x2 = t
$$\frac { d^{ 2 }y }{ dx^{ 2 } }$$ = $$\frac{d}{dx}$$ t-1/2
= $$\frac{d}{dt}$$ t-1/2 $$\frac{dt}{dx}$$
= – $$\frac{1}{2}$$ t-1/2-1 $$\frac{d}{dx}$$ (1 – x2)

Question 3.
If y = tan x + sec x then prove that:
$$\frac { d^{ 2 }y }{ dx^{ 2 } }$$ = $$\frac { cosx }{ (1-sinx)^{ 2 } }$$?
Solution:
y = tan x + sec x (given)
$$\frac{dy}{dx}$$ = sec2 x + sec x tan x
⇒ $$\frac{dy}{dx}$$ = sec x(sec x + tan x)
⇒ $$\frac{dy}{dx}$$ = $$\frac{1}{cosx}$$ [ $$\frac{1}{cosx}$$ + $$\frac{sinx}{cosx}$$ ]
= $$\frac { 1+sinx }{ cos^{ 2 }x }$$ = $$\frac { 1+sinx }{ 1-sin^{ 2 }x }$$
= $$\frac { 1+sinx }{ (1+sinx)(1-sinx) }$$
⇒ $$\frac{dy}{dx}$$ = $$\frac{1}{1-sinx}$$
Again differentiating both sides with respect to x,
$$\frac{d}{dx}$$ ( $$\frac{dy}{dx}$$ ) = $$\frac{d}{dx}$$ ( $$\frac{1}{1-sinx}$$
$$\frac { d^{ 2 }y }{ dx^{ 2 } }$$ = $$\frac { (1-sinx).0-1.(0-cosx) }{ (1-sinx)^{ 2 } }$$
⇒ $$\frac { d^{ 2 }y }{ dx^{ 2 } }$$ = $$\frac { cosx }{ (1-sinx)^{ 2 } }$$

Question 4.
If y = sin(sinx) then prove that:
y2 + y1 tan x + y cos2 x = 0? (CBSE 2018)
Solution:
y = sin(sin x)
Differentiating w.r.t. x,
y2 = cos (sinx) $$\frac{d}{dx}$$ (cos x) + cos x $$\frac{d}{dx}$$ {cos (sin x)}
= cos (sin x) (- sinx) + (cos x) [-sin(sin x)] cos x
⇒ y2 = – sin x cos (sin x) – cos2 x sin (sin x)
⇒ y2 = -sin x cos(sin x) – y cos2 x, [from eqn.(1)]
⇒ y2 = [ $$-\frac { sinx }{ cosx }$$. cos x] cos(sin x) – y cos2 x
⇒ y2 = – tan x {cos (sin x) cos x} – y cos2
⇒ y2 = -tan x {cos(sin x) cos x} – y cos2 x [from eqn.(2)]
⇒ y2 = (-tan x) y1 – y cos2 x, Proved.
⇒ y2 + y1 tan x + y cos2 x = 0.

Question 5.
If (x2 + y2)2 = xy then find $$\frac{dy}{dx}$$? (CBSE 2018)
Solution:
(x2 + y2)2 = xy
Differentiating with respect to x,
2(x2 + y2) (2x + 2y $$\frac{dy}{dx}$$ ) = x $$\frac{dy}{dx}$$ + y.1
⇒ 2(x2 + y2). 2x + 2(x2 + y2). 2y $$\frac{dy}{dx}$$ = x $$\frac{dy}{dx}$$ + y
⇒ [4y(x2 + y2) – x] $$\frac{dy}{dx}$$ = y – 4x (x2 + y2)
⇒ $$\frac{dy}{dx}$$ = $$\frac { y-4x(x^{ 2 }+y^{ 2 }) }{ 4(x^{ 2 }+y^{ 2 })y-x }$$

Question 6.
If y = 500e7x + 600e-7x then prove that:
$$\frac { d^{ 2 }y }{ dx^{ 2 } }$$ = 49 y? (NCERT)
Solution:
Given:
y = 500e7x + 600e-7x …………………….. (1)

Question 7.
If y = (tan-1 x)2 then prove that:
(x2 + 1)2 y2 + 2x (x2 + 1) y1 = 2? (NCERT)
Solution:
Given:
y = (tan-1 x)2

Question 8.
Differentiate sec-1 ( $$\frac { 1 }{ 2x^{ 2 }-1 }$$ ) with respect to: $$\sqrt { x^{ 2 }-1 }$$?
Solution:
Let y1 = sec-1 ( $$\frac { 1 }{ 2x^{ 2 }-1 }$$ )
⇒ y1 = cos-1 (2x2 – 1)

Question 9.
Differentiate tan-1 ( $$\frac { 2x }{ 1-x^{ 2 } }$$ ) with respect to:
sin-1 ( $$\frac { 2x }{ 1-x^{ 2 } }$$ )
Solution:
Let y1 = tan-1 $$\frac { 2x }{ 1+x^{ 2 } }$$ and y2 = sin-1 $$\frac { 2x }{ 1+x^{ 2 } }$$
Let x = tan θ, then θ = tan-1 x

⇒ y1 = tan-1 (tan 2θ) and y2 = sin-1 (sin 2θ)
⇒ y1 = 2θ and y2 = 2θ
⇒ y1 = 2 tan-1 x and y2 = 2 tan-1 x

Question 10.
Differentiate tan-1 ( $$\frac { \sqrt { 1+x^{ 2 }-1 } }{ x }$$ ) with respect to x?
Solution:
Let y1 = tan-1 ( $$\frac { \sqrt { 1+x^{ 2 }-1 } }{ x }$$ )
Put x = tan θ,

Question 11.
If x $$\sqrt { 1+y }$$ + y $$\sqrt { 1+x }$$ = 0 then prove that:
$$\frac{dy}{dx}$$ = -(1 + x)-2
Solution:
Given:
x$$\sqrt { 1+y }$$ + y $$\sqrt { 1+x }$$ = 0
⇒ x $$\sqrt { 1+y }$$ = -y$$\sqrt { 1+x }$$
Squaring both sides,
x2 (1 + y) = y2 (1 + x)
⇒ x2 + x2y = xy2 + y2
⇒ x2 – y2 + x2y – xy2 = 0
⇒ (x – y) (x + y) + xy (x – y) = 0
⇒ (x – y)(x + y + xy) = 0
⇒ x – y = 0
⇒ x = y
But x ≠ y
∴ x + y + xy = 0
⇒ y (l + x) = – x
∴ y = – $$\frac{x}{1+x}$$
Differentiating both sides with respect to x,

Question 12.
If xy = ey-x then prove that:
$$\frac{dy}{dx}$$ = $$\frac { 2-log_{ e }x }{ (1-log_{ e }x)^{ 2 } }$$
Solution:
Given: xy = ey-x
Applying log on both sides,
∴ loge xy = loge(ey-x)
⇒ y loge x = (y – x) loge e
⇒ y loge x – y = -x
⇒ y(1 – loge x) = x
⇒ y = $$\frac { x }{ 1-log_{ e }x }$$
Differentiating with respect to x,
Again,

Question 13.
If y$$\sqrt { 1-x^{ 2 } }$$ + x $$\sqrt { 1-y^{ 2 } }$$ then prove that:
$$\frac{dy}{dx}$$ + $$\sqrt{\frac{1-y^{2}}{1-x^{2}}}$$ = 0?
Solution:
Given:
y$$\sqrt { 1-x^{ 2 } }$$ + x $$\sqrt { 1-y^{ 2 } }$$ = 1.
Let x = sin θ and y = sin ϕ,

Question 14.
(A) If y = xsin-1x + xx then find the value of $$\frac{dy}{dx}$$?
Solution:
Given:
y = xsin-1x + xx
y = u + v
∴$$\frac{dy}{dx}$$ = $$\frac{du}{dx}$$ + $$\frac{dv}{dx}$$ ……………………. (1)
Where, u = xsin-1x
∴ log u = sin-1 x log x, (taking log both sides)
Differentiating both sides with respect to x,

and v = xx
∴ log v = x log x
Differentiating both sides with respect to x,
$$\frac{1}{v}$$. $$\frac{dv}{dx}$$ = 1.log x + x. $$\frac{1}{x}$$
⇒ $$\frac{dv}{dx}$$ = v(log x + 1)
⇒ $$\frac{dv}{dx}$$ = xx (log x + 1) ………… (3)∴ From eqn.(1),

(B) If y = x-1x + xx, then find the value of $$\frac{dy}{dx}$$?
Solution:
Solve like Q.No. 14(A).

Question 15.
If sin y = x sin (a + y) then prove that:
$$\frac{dy}{dx}$$ = $$\frac { sin^{ 2 }(a+y) }{ sina }$$?
Solution:
Given:
sin y = x sin (a + y)
⇒ x = $$\frac { siny }{ sin(a+y) }$$
Differentiating with respect to x,

Question 16.
If xy = ex-y then prove that:
$$\frac{dy}{dx}$$ = $$\frac { logx }{ (1+logx)^{ 2 } }$$?
Solution:
Given: xy = ex-y
Applying log on both sides,
y log x = (x – y) logea
⇒ y log x = (x – y).1 = x – y
Differentiating both sides with respect to x,

From eqn.(1),
y log x = x – y
⇒ y log x + y = x
⇒ y(logx + 1) = x
Put the value of y in eqn.(2)