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

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

Answers (1)

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.

 

Posted by

infoexpert26

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