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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

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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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