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  • Explain solution RD Sharma class 12 chapter 10 Differentiation exercise Multiple choice question 32 maths

Explain solution RD Sharma class 12 chapter 10 Differentiation exercise Multiple choice question 32 maths

Answers (1)

Answer:

        1

Hint:

        Differentiate the function w.r.t x

Given:

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

Solution: 

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

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

\frac{d y}{d x}=\frac{(\cos x-\sin x)^{2}}{(\cos x-\sin x)^{2}+(\sin x+\cos x)^{2}} \times  \left[\frac{(\cos x-\sin x) \frac{d}{d x}(\sin x+\cos x)-(\sin x+\cos x) \frac{d}{d x}(\cos x-\sin x)}{(\cos x-\sin x)^{2}}\right]

        =\frac{(\cos x-\sin x)^{2}}{(\cos x-\sin x)^{2}+(\sin x+\cos x)^{2}}\left[\frac{(\cos x-\sin x)(\cos x-\sin x)-(\sin x+\cos x)(-\sin x-\cos x)}{(\cos x-\sin x)^{2}}\right]

       =\frac{(\cos x-\sin x)^{2}}{(\cos x-\sin x)^{2}+(\sin x+\cos x)^{2}}\left[\frac{(\cos x-\sin x)(\cos x-\sin x)+(\sin x+\cos x)(\sin x+\cos x)}{(\cos x-\sin x)^{2}}\right]

      \begin{aligned} &=\frac{(\cos x-\sin x)^{2}}{(\cos x-\sin x)^{2}(\sin x+\cos x)^{2}} \times \frac{(\cos x-\sin x)^{2}+(\sin x+\cos x)^{2}}{(\cos x-\sin x)^{2}} \\ &\frac{d y}{d x}=1 \end{aligned}

    

 

 

 

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