#### Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 18 sub question (iv) maths textbook solution

Answer: $(x \cos x)^{x}\left\{(1-x \tan x)+\log (x \cos x)+(x \sin x)^{\frac{1}{x}}\left\{\frac{x \cos x}{x^{2} \sin x}-\log \left(\frac{x \sin x}{x^{2}}\right)\right\}\right.$

Hint: Diff by applying  $x^{x}$

Given: $(x \cos x)^{x}+(x \sin x)^{\frac{1}{x}}$

Solution:  $\text { Let } y=(x \cos x)^{x}+(x \sin x)^{\frac{1}{x}}$

$u=(x \cos x)^{x}$

Take log both sides

$\log u=\log (x \cos x)^{x}$

$\frac{1}{u} \times \frac{d u}{d x}=(x \cos x)^{x}\left[(\log x+1)+\left\{\log \cos x+\frac{x}{\cos x}(-\sin x)\right\}\right]$

$\frac{1}{u} \frac{d u}{d x}=(x \cos x)^{x}\{(1-x \tan x)+\log (x \cos x)\}$.....................(1)

$v=(x \sin x)^{\frac{1}{x}}$

Take log both side

$\log v=\frac{1}{x} \log (x \sin x)$

$\frac{1}{v} \frac{d v}{d x}=\frac{1}{x}\left\{\frac{1}{x \sin x}(x \cos x+\sin x \times 1\}+\log (x \sin x) \times \frac{-1}{x^{2}}\right.$

$\frac{1}{x}\left(\frac{x \cos x+\sin x \times 1}{x \sin x}\right)+\log (x \sin x) \times \frac{-1}{x^{2}}$

$\frac{d u}{d x}=(x \sin x)^{\frac{1}{x}}\left\{\frac{x \cos x+\sin x}{x^{2} \sin x}-\log \frac{(x \sin x)}{x^{2}}\right\}$        ............(2)

From (1) and (2)

$\frac{d y}{d x}=(x \cos x)^{x}\left\{(1-x \tan x)+\log (x \cos x)+(x \sin x)^{\frac{1}{x}}\left\{\frac{x \cos x}{x^{2} \sin x}-\log \left(\frac{x \sin x}{x^{2}}\right)\right\}\right.$

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