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  • Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 53 maths textbook solution

Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 53 maths textbook solution

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Answer:   \frac{d y}{d x}=(\sin x-\cos x)^{\sin x-\cos x}(\sin x+\cos x) \cdot(\log \sin x-\cos x)+1

Hint:  To solve this we take log both sides

Given: y=(\sin x-\cos x)^{\sin x-\cos x}

Solution:  

        y=(\sin x-\cos x)^{\sin x-\cos x}

Taking log on both sides,

        \log y=\log (\sin x-\cos x)^{\sin x-\cos x}                                    \left[\because \log x^{n}=n \log x\right]

 

        \log y=(\sin x-\cos x) \log (\sin x-\cos x)

 

        \frac{1}{y} \frac{d y}{d x}=\log (\sin x-\cos x)(\cos x+\sin x)+(\sin x-\cos x)\left(\frac{1}{(\sin x-\cos x)}(\cos x+\sin x)\right)

 

        \begin{array}{r} \frac{1}{y} \frac{d y}{d x}=(\cos x+\sin x)[1+\log (\sin x-\cos x)] \frac{d y}{d x}=y(\sin x+\cos x)[\log (\sin x- \cos x)+1 \end{array}

     

        \frac{d y}{d x}=(\sin x-\cos x)^{\sin x-\cos x}(\sin x+\cos x) \cdot(\log \sin x-\cos x)+1

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