ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.8

Utilizing Understanding ISC Mathematics Class 12 Solutions Chapter 5 Continuity and Differentiability Ex 5.8 as a study aid can enhance exam preparation.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.8

Differentiate the following (1 to 6) functions w.r.t. x:

Question 1.
(i) e^x + 3 sin x
(ii) 10^x + \(\frac{1}{3}\) e^x – 2 log x
Solution:
(i) Let y = e^x + 3 sin x ;
Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = e^x + 3 cos x

(ii) Let y = 10^x + \(\frac{1}{3}\) e^x – 2 log x
Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = 10^x log 10 + \(\frac{1}{3}\) e^x – \(\frac{2}{x}\)

Question 2.
(i) e^-x (NCERT)
(ii) sin (log x), x > 0 (NCERT)
Solution:
(i) Let y = e^-x ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = – e^-x

(ii) Let y = sin (log x), x > 0
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = cos (log x) \(\frac{d}{d x}\) log x
= \(\frac{\cos (\log x)}{x}\).

Question 3.
(i) e^{cos x} (NCERT)
(ii) e^{sin-1 x} (NCERT)
Solution:
(i) Let y = e^{cos x}
On differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = e^{cos x} \(\frac{d}{d x}\) (cos x)
= e^{cos x} (- sin x)

(ii) Let y = e^{sin-1 x}
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = e^{sin-1 x} \(\frac{d}{d x}\) (sin^-1 x)
= e^{sin-1 x} \(\frac{1}{\sqrt{1-x^2}}\)

Question 4.
(i) e^x3 (NCERT)
(ii) \(\sqrt{e^{\sqrt{x}}}\), x > 0
Solution:
(i) Let y = e^x3 ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = e^x3 \(\frac{d}{d x}\) x³
= e^x3 . 3x²

(ii) Let y = \(\sqrt{e^{\sqrt{x}}}\), x > 0
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{d}{d x}\left(e^{\sqrt{x}}\right)^{1 / 2}\)
= \(\frac{1}{2}\left(e^{\sqrt{x}}\right)^{\frac{1}{2}-1} \frac{d}{d x} e^{\sqrt{x}}\)
= \(\frac{1}{2 \sqrt{e^{\sqrt{x}}}} e^{\sqrt{x}} \frac{1}{2 \sqrt{x}}\)
= \(\frac{e^{\sqrt{x}}}{4 \sqrt{x} \sqrt{e^{\sqrt{x}}}}\)

Question 5.
(i) log (log x), x > 1 (NCERT)
(ii) log₇ (log x), x > 1 (NCERT)
Solution:
(i) Let y = log (log x), x > 1
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) log (log x)
= \(\frac{1}{\log x}\) \(\frac{d}{d x}\) (log x)
= \(\frac{1}{x \log x}\)

(ii) Let y = log₇ (log x)
Differentiating both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\log (\log x)}{\log 7}\right)\)
= \(\frac{1}{\log 7} \frac{1}{\log x} \frac{1}{x}\)
= \(\frac{1}{x \log x \log 7}\)

Question 6.
(i) log (cos e^x) (NCERT)
(ii) tan^-1 (e^{sin x})
Solution:
(i) Let y = log (cos e^x) ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{1}{\cos e^x}\) \(\frac{d}{d x}\) cos e^x
= \(\frac{1}{\cos e^x}\) (- sin e^x) \(\frac{d}{d x}\) e^x
= – e^x tan (e^x)

(ii) Let y = tan^-1 (e^{sin x})
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{1}{1+\left(e^{\sin x}\right)^2} \frac{d}{d x}\) e^{sin x}
= \(\frac{1}{1+e^{2 \sin x}}\) e^{sin x} . cos x

Question 7.
Is the function f(x) = log (2x – 1) derivable at x = 0?
Solution:
(i) Given f(x) = log (2x – 1)
for domain of f : f(x) must be a real number
⇒ log (2x – 1) must be a real number
⇒ 2x – 1 > 0
⇒ x > \(\frac{1}{2}\)
∴ D_f = (\(\frac{1}{2}\), ∞).

Question 8.
If f(x) = log (log x), find f'(e).
Solution:
Given f(x) = log (log x) ;
Diff. both sides w.r.t. x, we have
f'(x) = \(\frac{1}{\log x} \cdot \frac{1}{x}\)
∴ f'(e) = \(\frac{1}{\log e} \cdot \frac{1}{e}\)
= \(\frac{1}{e \log e}\)
= \(\frac{1}{e \times 1}=\frac{1}{e}\)

Question 9.
If y = log (tan x), show that \(\frac{d y}{d x}\) = 2 cosec 2x.
Solution:
Let y = log (tan x) ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=-\frac{1}{\tan x}\) sec² x
= \(\frac{1}{\frac{\sin x}{\cos x} \times \cos ^2 x}\)
= \(\frac{2}{\sin 2 x}\)
\(\frac{d y}{d x}\) = 2 cosec 2x.

Question 10.
If f(1) = 4 and f'(1) = 2, find the value of the derivative of log f(e^x) w.r.t. x at x = 0.
Solution:
Let y = log f(e^x)
Differentiating w.r.t. x, we have
\(\frac{d y}{d x}=\frac{1}{f\left(e^x\right)} \frac{d}{d x}\) f(e^x)
= \(\frac{f^{\prime}\left(e^x\right)}{f\left(e^x\right)}\) × e^x
at x = 0,
\(\frac{d y}{d x}=\frac{f^{\prime}(1)}{f(1)}\) × e⁰
= \(\frac{2}{4}=\frac{1}{2}\)
[Since f'(1) = 2 and f(1) = 4]

Question 11.
If f(x) = e^x g(x), g(0) = 2 and g'(0) = 1, then find f'(0).
Solution:
Given f(x) = e^x g(x)
Diff. both sides w.r.t. x, we have
f'(x) = e^x g'(x) + g(x) e^x …………….(1)
putting x = 0 in eqn. (1) ; we have
f'(0) = e⁰ g'(0) + g(0) e⁰
= g'(0) + g(0)
= 1 + 2 = 3
[∵ g(0) = 2 and g'(0) = 1]

Differentiate the following functions w.r.t. x :

Question 12.
(i) \(\frac{\mathcal{e}^x}{\sin x}\)
(ii) \(\frac{e^x}{1+\sin x}\).
Solution:
(i) Let y = \(\frac{\mathcal{e}^x}{\sin x}\)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{\sin x \frac{d}{d x} e^x \frac{d}{d x} \sin x}{\sin ^2 x}\)
[using quotient rule]
= \(\frac{(\sin x) e^x-e^x \cos x}{\sin ^2 x}\)
= \(\frac{e^x(\sin x-\cos x)}{\sin ^2 x}\)

(ii) Given y = \(\frac{e^x}{1+\sin x}\) ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{(1+\sin x) e^x-e^x \cos x}{(1+\sin x)^2}\)
= \(\frac{e^x(1+\sin x-\cos x)}{(1+\sin x)^2}\)

Question 13.
(i) 2x tan^-1 x – log (1 + x²)
(ii) \(\frac{\cos x}{\log x}\), x > 0 (NCERT)
Solution:
(i) Let y = 2x tan^-1 x – log (1 + x²)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = 2 [x . \(\frac{1}{1+x^2}\) + tan^-1 x . 1] – \(\frac{2 x}{1+x^2}\)
= 2 tan^-1 x

(ii) Let y = \(\frac{\cos x}{\log x}\), x > 0
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{\log x \frac{d}{d x} \cos x-\cos x \frac{d}{d x} \log x}{(\log x)^2}\) [quotient rule]
= \(\frac{-\sin x \log x-\frac{\cos x}{x}}{(\log x)^2}\)
= \(\frac{-x \sin x \log x+\cos x}{x(\log x)^2}\)

Question 14.
(i) e^{sec x2} + 3 cos^-1 x (NCERT)
(ii) e^x + e^x2 + ………….. + e^x5.
Solution:
(i) Let y = e^{sec x2} + 3 cos^-1 x
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = e^{sec x2} \(\frac{d}{d x}\) sec² x + \(\frac{d}{d x}\) 3 cos^-1 x
= e^{sec x2} (2 sec x) \(\frac{d}{d x}\) sec x + 3 \(\left(\frac{-1}{\sqrt{1-x^2}}\right)\)
= 2 sec² x tan x e^{sec2 x} – \(\frac{3}{\sqrt{1-x^2}}\)

(ii) Let y = e^x + e^x2 + ………….. + e^x5
i.e. y = e^x + e^x2 + e^x3 + e^x4 + e^x5
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = e^x + e^x2 \(\frac{d}{d x}\) x² + e^x3 \(\frac{d}{d x}\) x³ + e^x4 \(\frac{d}{d x}\) x⁴ + e^x5 \(\frac{d}{d x}\) x⁵
= e^x + 2x e^x2 + 3x² e^x3 + 4x³ e^x4 + 5x⁴ e^x5

Question 15.
(i) \(\frac{5^x \log x}{x^2+1}\)
(ii) cos (log x + e^x). (NCERT)
Solution:
(i) Let y = \(\frac{5^x \log x}{x^2+1}\)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{\left(x^2+1\right) \frac{d}{d x}\left(5^x \log x\right)-\left(5^x \log x\right) \frac{d}{d x}\left(x^2+1\right)}{\left(x^2+1\right)^2}\)
= \(\frac{\left(x^2+1\right)\left(\frac{5^x}{x}+\log x \cdot 5^x \log 5\right)-\left(5^x \log x\right) 2 x}{\left(x^2+1\right)^2}\)
= \(\frac{5^x\left[\left(x^2+1\right)(1+x \log 5 \cdot \log x)-2 x^2 \log x\right]}{x\left(x^2+1\right)^2}\)

(ii) Let y = cos (log x + e^x)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = – sin (log x + e^x) . \(\frac{d}{d x}\) (log x + e^x)
= – sin (log x + e^x) (\(\frac{1}{x}\) + e^x)
∴ \(\frac{d y}{d x}\) = – \(\frac{1}{x}\) sin (log x + e^x) (1 + x e^x).

Question 16.
(i) log \(\left(\frac{x+\sqrt{x^2-a^2}}{x-\sqrt{x^2-a^2}}\right)\)
(ii) log \(\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\)
Solution:
(i) Let y = log \(\left(\frac{x+\sqrt{x^2-a^2}}{x-\sqrt{x^2-a^2}}\right)\)
Diff. both sides w.r.t. x, we have

(ii) y = log \(\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\)
= \(\frac{1}{2}\) [log (1 – cos x) – log (1 + cos x)]
[∵ log a^b = b log a ;
log \(\frac{a}{b}\) = log a – log b]
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{1}{2}\left[\begin{array}{c}
\frac{1}{1-\cos x}(+\sin x) \\
-\frac{1}{1+\cos x}(-\sin x)
\end{array}\right]\)
= \(\frac{1}{2}\left[\frac{\sin x(1+\cos x+1-\cos x)}{(1-\cos x)(1+\cos x)}\right]\)
= \(\frac{\sin x}{1-\cos ^2 x}\)
= \(\frac{\sin x}{\sin ^2 x}\)
= cosec x.

Question 17.
(i) e^{cot-1 x2}
(ii) x \(\sqrt{1+x^2}\) + log (x + \(\sqrt{x^2+1}\)).
Solution:
(i) Let y = e^{cot-1 x2} ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = e^{cot-1 x2} \(\frac{d}{d x}\) cot^-1 x²
= e^{cot-1 x2} \(\left\{\frac{-1}{1+\left(x^2\right)^2} \times 2 x\right\}\)
= – \(\frac{2 x e^{\cot ^{-1} x^2}}{1+x^4}\)

(ii) Let y = x \(\sqrt{1+x^2}\) + log (x + \(\sqrt{x^2+1}\))
Diff. both sides w.r.t. x, we get

Question 18.
(i) If y = \(\frac{\log \left(x+\sqrt{x^2+1}\right)}{\sqrt{x^2+1}}\), prove that (x² + 1) \(\frac{d y}{d x}\) + xy = 1.
(ii) If y = e^{2 log x + 3x}, prove that \(\frac{d y}{d x}\) = x (2 + 3x) e^3x.
Solution:
(i) Given y \(\sqrt{x^2+1}\) = log (\(\sqrt{x^2+1}\) + x)
Differentiating both sides w.r.t. x ; we get

⇒ xy + (x² + 1) \(\frac{d y}{d x}\) = 1
⇒ (x² + 1) \(\frac{d y}{d x}\) + xy = 1.

(ii) Let y = e^{2 log x + 3x}
= e^{2 log x} . e^3x
= e^{log x2} e^3x
⇒ y = x² e^3x . 3 + e^3x . 2x
= e^3x (2 + 3x) x.

Question 19.
Differentiate tan^-1 \(\left(\frac{2^{x+1}}{1-4^x}\right)\) w.r.t. x.
Solution:
Let y = tan^-1 \(\left(\frac{2^{x+1}}{1-4^x}\right)\)
= tan^-1 \(\left(\frac{2 \times 2^x}{1-\left(2^x\right)^2}\right)\)
put 2^x = tan θ
i.e. θ = tan^-1 (2x)
Then y = tan^-1 \(\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right)\)
⇒ y = tan^-1 (tan 2θ) = 2θ
⇒ y = 2 tan^-1 (2^x) ;
Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}=\frac{2}{1+\left(2^x\right)^2} \frac{d}{d x}\) (2^x)
= \(\frac{2}{1+2^{2 x}}\) 2^x log 2
= \(\frac{2^{x+1} \log 2}{1+4^x}\).

Question 20.
Find \(\frac{d y}{d x}\) when
(i) xy + xe^-y + ye^x = x²
(ii) e^x-y = log \(\left(\frac{x}{y}\right)\)
Solution:
(i) Given, xy + xe^-y + ye^x = x² …………..(1)
Diff. eqn. (1) w.r.t. x, taking y as a function of x, we have
x \(\frac{d}{d x}\) + y . 1 + x e^-y (- \(\frac{d y}{d x}\)) + e^-y + y e^x + e^x \(\frac{d y}{d x}\) = 2x
⇒ [x – xe^-y + e^x] \(\frac{d y}{d x}\) = 2x – y e^x – e^-y – y
∴ \(\frac{d y}{d x}\) = \(\frac{2 x-y e^x-e^{-y}-y}{x+e^x-x e^{-y}}\)

(ii) Given e^x-y = log \(\left(\frac{x}{y}\right)\)
Diff. eqn. (1) both sides w.r.t. x, we have
\(e^{x-y}\left[1-\frac{d y}{d x}\right]=\frac{1}{x}-\frac{1}{y} \frac{d y}{d x}\)
⇒ \(\left[\frac{1}{y}-e^{x-y}\right] \frac{d y}{d x}=\frac{1}{x}-e^{x-y}\)
⇒ \(\frac{\left[1-y e^{x-y}\right]}{y} \frac{d y}{d x}=\left[\frac{1-x e^{x-y}}{x}\right]\)
⇒ \(\frac{d y}{d x}=\frac{y\left(1-x e^{x-y}\right)}{x\left(1-y e^{x-y}\right)}\).

Question 21.
If log (x² + y² = 2 tan^-1 \(\frac{y}{x}\), show that \(\frac{d y}{d x}=\frac{x+y}{x-y}\).
Solution:
Given, log (x² + y² = 2 tan^-1 \(\frac{y}{x}\)
diff. both sides w.r.t. x ; we have
\(\frac{1}{x^2+y^2}\left[2 x+2 y \frac{d y}{d x}\right]=\frac{2}{1+\frac{y^2}{x^2}}\left[\frac{x \frac{d y}{d x}-y \cdot 1}{x^2}\right]\)
⇒ \(\frac{1}{x^2+y^2}\left[x+y \frac{d y}{d x}\right]=\frac{1}{x^2+y^2}\left[x \frac{d y}{d x}-y\right]\)
⇒ (y – x) \(\frac{d y}{d x}\) = – x – y
⇒ \(\frac{d y}{d x}\) = \(\frac{x+y}{x-y}\).

Question 22.
If y log x = x – y, prove that \(\frac{d y}{d x}\) = \(\frac{\log x}{(1+\log x)^2}\).
Solution:
Given y log x = x – y
⇒ y (1 + log x) = x
⇒ y = \(\frac{x}{1+\log x}\) ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{(1+\log x) \frac{d}{d x} x-x \frac{d}{d x}(1+\log x)}{(1+\log x)^2}\)
\(\frac{d y}{d x}=\frac{(1+\log x) \cdot 1-x\left(0+\frac{1}{x}\right)}{(1+\log x)^2}\)
= \(\frac{1+\log x-1}{(1+\log x)^2}\)
⇒ \(\frac{d y}{d x}=\frac{\log x}{(1+\log x)^2}\).

Question 23.
Differentiate the following functions w.r.t. x :
(i) log_x (2x – 3)
(ii) log_{cos x} sin x.
Solution:
(i) Let y = log_x (2x – 3)
= \(\frac{\log (2 x-3)}{\log x}\)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{\left[(\log x) \frac{d}{d x} \log (2 x-3)-\log (2 x-3) \frac{d}{d x} \log x\right]}{(\log x)^2}\)
= \(\frac{\left[\log x \cdot \frac{2}{2 x-3}-\frac{\log (2 x-3)}{x}\right]}{(\log x)^2}\)
= \(\left[\frac{2 x \log x-(2 x-3) \log (2 x-3)}{(2 x-3) x(\log x)^2}\right]\).

(ii) Let y = log_{cos x} sin x
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{\log \cos x \frac{d}{d x}(\log \sin x)-\log \sin x \frac{d}{d x} \log \cos x}{[\log \cos x]^2}\)
= \(\frac{\log \cos x \cdot \frac{\cos x}{\sin x}-(\log \sin x) \cdot\left(-\frac{\sin x}{\cos x}\right)}{(\log \cos x)^2}\)
= \(\frac{\cot x \log \cos x+\tan x \log \sin x}{(\log \cos x)^2}\).