OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(k)

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S Chand Class 12 ICSE Maths Solutions Chapter 8 Differentiation Ex 8(k)

Find the derivative of the following functions :

Question 1.
(x² + 2)³ (1 – x³)⁴
Solution:
Let y – (x² + 2)³ (1 – x³)⁴
Taking logarithm on both sides, we have
log y – log (x² + 2)³ (1 – x³)⁴
⇒ log y = 3 log(x² + 2) + 4log(1 – x³)
[∵ log ab = log a + log b and log a^b = b log a]
Diff. both sides w.r.t. x; we have

Question 2.
\(\frac{x\left(1-x^2\right)^2}{\left(1+x^2\right)^{1 / 2}}\)
Solution:
Let y = \(\frac{x\left(1-x^2\right)^2}{\left(1+x^2\right)^{1 / 2}}\)
Taking logarithm on both sides, we have

Question 3.
\(\frac{(x+1)^2 \sqrt{(x-1)}}{(x+4)^3 e^x}\)
Solution:

Question 4.
\(\sqrt{(x-1)(x-2)(x-3)(x-4)}\)
Solution:

Question 5.
\(\frac{(x-a)(x-b)}{(x-p)(x-q)}\)
Solution:

Question 6.
\(\frac{2(x-\sin x)^{3 / 2}}{\sqrt{x}}\)
Solution:
Let y = \(\frac{2(x-\sin x)^{3 / 2}}{\sqrt{x}}\)
Taking logarithm on both sides, we have
log y = log 2 + \(\frac { 3 }{ 2 }\)log (x – sin x) – \(\frac { 1 }{ 2 }\) log x
Diff. both sides w.r.t. x; we have
\(\frac{1}{y} \frac{d y}{d x}=\left[\frac{3}{2} \frac{1}{x-\sin x}[1-\cos x]-\frac{1}{2 x}\right]\)
Thus, \(\frac{d y}{d x}=\frac{2(x-\sin x)^{3 / 2}}{\sqrt{x}}\) \(\left[\frac{3}{2}\left(\frac{1-\cos x}{x-\sin x}\right)-\frac{1}{2 x}\right]\)

Question 7.
(i) x^1/x
(ii) \(x^{\sqrt{x}}\)
(iii) \(\left(\frac{1}{x}\right)^x\)
Solution:
(i) Let y = x^1/x ;
Taking logarithm on both sides, we have
log y = \(\frac { 1 }{ x }\) log x ;
DifF. both sides w.r.t. x; we have
\(\frac{1}{y} \frac{d y}{d x}=\frac{1}{x} \cdot \frac{1}{x}+\log x\left(-\frac{1}{x^2}\right)\)
⇒ \(\frac{d y}{d x}=\frac{x^{1 / x}}{x^2}(1-\log x)\)

(ii) Let y = \(x^{\sqrt{x}}\) ;
Taking logarithm on both sides, we have
log y = log \(x^{\sqrt{x}}\) = \(\sqrt{x}\) log x ;
Diff. both sides w.r.t. x; we have
\(\frac{1}{y} \frac{d y}{d x}=\frac{\sqrt{x}}{x}+\log x \frac{1}{2 \sqrt{x}}\) ;
⇒ \(\frac{d y}{d x}=y\left[\frac{1}{\sqrt{x}}+\frac{\log x}{2 \sqrt{x}}\right]\)
= \(x^{\sqrt{x}}\left[\frac{1}{\sqrt{x}}+\frac{\log x}{2 \sqrt{x}}\right]\)

(iii) \(\left(\frac{1}{x}\right)^x\) ;
Taking logarithm on both sides, we have
log y = log \(\left(\frac{1}{x}\right)^x\) = x log (\(\frac { 1 }{ x }\))
= x(- log x) = – x log x
Diff. both sides w.r.t. x; we have
\(\frac{1}{y} \frac{d y}{d x}=-\left[x \cdot \frac{1}{x}+\log x \cdot 1\right]=-(1+\log x)\) ;
⇒ \(\frac{d y}{d x}\) = – y(1 + log x)
= – \(\left(\frac{1}{x}\right)^x\) (1 + log x)

Question 8.
(sin x)^x
Solution:
Let y = (sin x)^x;
Taking logarithm on both sides, we have
log y = x log sin x ;
Diff. both sides w.r.t. x; we have
\(\frac{1}{y} \frac{d y}{d x}=\frac{x}{\sin x}\) cos x + log sinx
⇒ \(\frac{d y}{d x}\) = y[xcotx + log sin x]
= (sin x)^x[x cot x + log sin x]

Question 9.
x^{sin x}
Solution:
Let y = x^{sin x} ;
Taking logarithm on both sides, we have
log y = log x^{sin x} = sin x . log x ;
Diff. both sides w.r.t. x; we have
\(\frac{1}{y} \frac{d y}{d x}=\frac{\sin x}{x}\) cos x + logxcosx
⇒ \(\frac{d y}{d x}\) = y\(\left[\frac{\sin x}{x}+(\log x) \cos x\right]\)
= x^{sin x}\(\left[\frac{\sin x}{x}+\cos x \log x\right]\)

Question 10.
(sin x)^{tan x}
Solution:
Let y = (sin x)^{tan x} ;
Taking logoritum on both sides, we have
log y = log (sin x)^{tan x}
= tan x . log sin x ;
Diff. both sides w.r.t. x; we have
\(\frac{1}{y} \frac{d y}{d x}=\tan x \frac{1}{\sin x}\) cos x + log sin x sec² x
⇒ \(\frac{d y}{d x}\) = y[1 + sec² x log sin x]
= (sin x)^{tan x}[1 + sec² x log sin x]

Question 11.
(tan x)^{log x}
Solution:
Let y = (tan x)^{log x} ;
Taking logarithm on both sides, we have
log y = log (tan x)^{log x} – log x . log tan x ;
Diff. both sides w.r.t. x; we have
\(\frac{1}{y} \frac{d y}{d x}=\log x \frac{1}{\tan x} \sec ^2 x+\log \tan x \cdot \frac{1}{x} \Rightarrow \frac{d y}{d x}=y\left[\frac{\log x}{\sin x \cos x}+\frac{\log \tan x}{x}\right]\)
⇒ \(\frac{d y}{d x}=(\tan x)^{\log x}\left[\frac{\log x}{\sin x \cos x}+\frac{\log \tan x}{x}\right]\)

Question 12.
x^{log x}
Solution:
Let y = x^{log x} ;
Taking logarithm on both sides, we have
log y = log x^{log x} = log x . log x = (log x)² ;
Diff. both sides w.r.t. x; we have
\(\frac{1}{y} \frac{d y}{d x}=2 \log x \cdot \frac{1}{x} \Rightarrow \frac{d y}{d x}=y\left[\frac{2 \log x}{x}\right]=x^{\log x} \cdot \frac{2 \log x}{x}\)

Question 13.
(tan x)^{cos x}
Solution:

Question 14.
(log x)^x
Solution:
Let y = (log x)^x ;
Taking logarithm on both sides, we have
log y = log(log x)^x = x log(log x) ;
Diff. both sides w.r.t. x; we have
\(\frac{1}{y} \frac{d y}{d x}=\log (\log x) \cdot 1+x \frac{1}{\log x} \cdot \frac{1}{x}\)
⇒ \(\frac{d y}{d x}=y\left[\log (\log x)+\frac{1}{\log x}\right] \Rightarrow \frac{d y}{d x}=(\log x)^x\left[\log (\log x)+\frac{1}{\log x}\right]\)

Question 15.
\(\left(1+\frac{1}{x}\right)^x\)
Solution:

Question 16.
\(x^x \sqrt{x}\)
Solution:

Question 17.
(i) cos x^x
(ii) sin (x^x)
(iii) If y = \(\left(\tan \frac{\pi x}{4}\right)^{\frac{4}{\pi x}}\), find \(\frac { dy }{ dx }\) at x = 1.
Solution:
(i) Let y = cos x^x ;
Diff. both sides w.r.t. x; we have

Question 18.
Find \(\frac { dy }{ dx }\) if
(i) y = log (x^x + cosec² x)
(ii) y = \(e^{\sin ^2 x}\left(2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right)\)
(ii) x^y = y^x
(iv) (cos x)^y = (sin y)^x
Solution:
(i) Given y = log (x^x + cosec² x); Diff both sides w.r.t. x; we have

(ii) Given y = \(e^{\sin ^2 x}\left(2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right)\)
Taking logarithm on both sides; we have

(iii) x^y = y^x;
Taking logarithm on both sides y log x = x log y ;
Diff. both sides w.r.t. x; we have
\(\frac{y}{x}+\log x \frac{d y}{d x}=\frac{x}{y} \frac{d y}{d x}\) + log y.1
⇒ \(\left(\log x-\frac{x}{y}\right) \frac{d y}{d x}=\log y-\frac{x}{y}\)
⇒ \(\frac{d y}{d x}=\frac{y(x \log y-y)}{x(y \log x-x)}\)

(iv) Given (cos x)^y = (sin y)^x;
Taking logarithm on both sides; we have
log (cos x)^y = log (sin y)^x
y log cos x = x log sin y;
Diff. both sides w.r.t. x; we have

Question 19.
IF y = e^x-y, show that \(\frac { 1 }{ 2 }\).
Solution:
Given y = e^x-y ;
Taking logarithm on both sides; we have
log y = log e^x-y = x – y ;
Diff. both sides w.r.t. x; we have
⇒ \(\frac{1}{y} \frac{d y}{d x}=1-\frac{d y}{d x}\)
⇒ \(\left(\frac{1}{y}+1\right) \frac{d y}{d x}\) = 1 ⇒ \(\left(\frac{1+y}{y}\right) \frac{d y}{d x}\) = 1
⇒ \(\frac{d y}{d x}=\frac{y}{1+y}\)

Question 20.
If x^myⁿ = (x + y)^m+n, prove that \(\frac{d y}{d x}=\frac{y}{x}\).
Solution:
Given x^myⁿ = (x + y)^m+n ;
Taking logarithm on both sides; we have
log x^m + log yⁿ = log (x + y)^m+n
⇒ m log x + n log y = (m + n) log(x + y) [∵ log ab = log a + log b & log a^b = b log a]
Diff. both sides w.r.t. x; we have

Question 21.
If y = \(x^{x^{r \ldots \ldots \ldots \infty}}\), prove that \(\frac{d y}{d x}=\frac{y^2}{x(1-y \log x)}\)
Solution:

Question 22.
If y = \(\sqrt{x}^{\sqrt{x}^{\sqrt{x} \ldots \ldots}}\) find \(\frac { dy }{ dx }\).
Solution:
Given y = \(\sqrt{x}^{\sqrt{x}^{\sqrt{x} \ldots \infty}}=(\sqrt{x})^y\)
Taking logarithm on both sides; we have

Question 23.
If y = \(a^{x^{a^{x^{a^x \ldots \ldots \ldots . . \infty}}}}\) find \(\frac { dy }{ dx }\).
Solution:
Given y = \(a^{x^y}\)
Taking logarithm on both sides; we have
log y = log \(a^{x^y}\) = log a A
gain taking logarithm on both sides, we have
log log y = log(x^y log a) = y log x + log log a;
Diff. both sides w.r.t. x; we have

Question 24.
If y = x^y, prove that x\(\frac{d y}{d x}=\frac{y^2}{(1-y \log x)}\).
Solution:
Given y = \(x^{x^{x \ldots \ldots \infty}}=x^y\)
Taking logarithm on both sides; we have
log y = y log x ;
Diff. both sides w.r.t. x; we have

Question 25.
Find \(\frac { dy }{ dx }\) when x^y + y^x = c.
Solution:
Given x^y + y^x = c ⇒ u + v = c …(1)
where u = x^y …(2)
& v = y^x …(3)
Diff. eqn. (1) both sides w.r.t. x; we get
\(\frac{d u}{d x}+\frac{d v}{d x}=0\) … (4)
Taking logarithm on both sides of eqn. (2); we have
log u = y log x; diff. w.r.t. x, we have
\(\frac{1}{u} \frac{d u}{d x}=\frac{y}{x}+\log x \frac{d y}{d x}\)
⇒ \(\frac{d u}{d x}=x^y\left[\frac{y}{x}+\log x \frac{d y}{d x}\right]\) … (5)
Taking logaritum on both sides of eqn.(3); we have
log v = x log y
Diff. both sides w.r.t. x; we have

Question 26.
If x^y = e^y-x, prove that \(\frac{d y}{d x}=\frac{2-\log x}{(1-\log x)^2}\).
Solution:
Given x^y = e^y-x;
Taking logarithm on both sides; we have
y log x = (y – x) log e – y – x
⇒ y(1 – log x) = x

Question 27.
Differentiate (sin x)^x w.r.t. x².
Solution:
Let y = (sin x)^x …(1)
& z = x² …(2)
Taking logarithm on both sides of eqn. (1); we have
log y = x log sin x ; diff. w.r.t. x, we have