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

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

Differentiation Important Questions

Differentiation Short Answer Type 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 e^x] with respect to x? (NCERT)
Solution:
Given:
y = log [cos e^x]
\(\frac { dy }{ dx } \) = \(\frac { d }{ dx } \) [log (cos e^x)]
\(\frac { dy }{ dx } \) = \(\frac { d }{ dx } \) log t, [Putting cos e^x = t]
= \(\frac { d }{ dt } \) log t \(\frac { dt }{ dx } \)
= \(\frac { 1 }{ t } \). \(\frac { d }{ dx } \) cos e^x
= \(\frac { 1 }{ cose^{ x } } \) × \(\frac { d }{ dx } \) cos u, [Putting e^x = u]
= \(\frac { 1 }{ cose^{ x } } \). \(\frac { d }{ du } \) cos u \(\frac { du }{ dx } \)
= \(\frac { -sinu }{ cose^{ x } } \). \(\frac { d }{ dx } \) e^x
= \(\frac { -(sine^{ x })e^{ x } }{ cose^{ x } } \)
= – e^x tan e^x

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

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

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 = e^msin-1x then prove that (1 – x²) y₂ – xy₁ – m²y = 0?
Solution:
Given:
y = e^msin-1x
\(\frac { dy }{ dx } \) = y₁ = e^msin-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 = e^mtan-1x then prove that (1 + x²) y₂ + (2x – m) y₁ = 0?
Solution:
Solve like Q.5 (A)

Question 5.
(C) If y = e^mcos-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
Answer:
\(\frac { dy }{ dx } \) = \(\frac { 1 }{ 2\sqrt { 1-x^{ 2 } } } \)

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 = x^sinx then find the value of \(\frac { dy }{ dx } \)?
Solution:
Given:
y = x^sinx
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 = e^x+y
Taking log on both sides,
log y = log e^x+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 2^x = tan θ
Then, θ = tan^-1 2^x
⇒ 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 (2^x), [θ = tan^-1(2^x)]
∴ \(\frac{dy}{dx}\) = 2 \(\frac{d}{dx}\) tan^-1 (2^x)
Putting 2^x = t
⇒ \(\frac{dy}{dx}\) = 2 \(\frac{d}{dx}\) tan^-1 t
= 2 \(\frac{d}{dt}\) tan^-1 t\(\frac{dt}{dx}\)
⇒ \(\frac{dy}{dx}\) = \(\frac { 2 }{ 1+t^{ 2 } } \) \(\frac{d}{dx}\) (2^x),
= \(\frac { 2 }{ 1+2^{ 2x } } \) × 2^x 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 = 500e^7x + 600e^-7x then prove that:
\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = 49 y? (NCERT)
Solution:
Given:
y = 500e^7x + 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 x^y = e^y-x then prove that:
\(\frac{dy}{dx}\) = \(\frac { 2-log_{ e }x }{ (1-log_{ e }x)^{ 2 } } \)
Solution:
Given: x^y = e^y-x
Applying log on both sides,
∴ log_e x^y = log_e(e^y-x)
⇒ y log_e x = (y – x) log_e e
⇒ y log_e x – y = -x
⇒ y(1 – log_e 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 = x^sin-1x + x^x then find the value of \(\frac{dy}{dx}\)?
Solution:
Given:
y = x^sin-1x + x^x
y = u + v
∴\(\frac{dy}{dx}\) = \(\frac{du}{dx}\) + \(\frac{dv}{dx}\) ……………………. (1)
Where, u = x^sin-1x
∴ log u = sin^-1 x log x, (taking log both sides)
Differentiating both sides with respect to x,


and v = x^x
∴ 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)
⇒ \(\frac{dv}{dx}\) = x^x (log x + 1) ………… (3)∴ From eqn.(1),

(B) If y = x^-1x + x^x, 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 x^y = e^x-y then prove that:
\(\frac{dy}{dx}\) = \(\frac { logx }{ (1+logx)^{ 2 } } \)?
Solution:
Given: x^y = e^x-y
Applying log on both sides,
y log x = (x – y) log_ea
⇒ 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)

MP Board Class 12 Maths Important Questions