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Please solve RD Sharma class 12 chapter 10 Differentiation exercise Multiple choice question 13 maths textbook solution

Answers (1)

Answer:

        -1

Hint:

        Differentiate the function w.r.t x

Given:

        \sin (x+y)=\log (x+y)

Solution:  

        \begin{aligned} &\sin (x+y)=\log (x+y) \\\\ &\cos (x+y)\left(1+\frac{d y}{d x}\right)=\frac{1}{(x+y)}\left(1+\frac{d y}{d x}\right) \end{aligned}

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

        \begin{aligned} &{\left[\cos (x+y)-\frac{1}{(x+y)}\right] \frac{d y}{d x}=\frac{1}{(x+y)}-\cos (x+y)} \\\\ &-\left[\frac{1}{(x+y)}-\cos (x+y)\right] \frac{d y}{d x}=\frac{1}{(x+y)}-\cos (x+y) \\\\ &\frac{d y}{d x}=-1 \end{aligned}

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