# Choose the correct alternative in the following:

Question:

Choose the correct alternative in the following:

If $\sin (x+y)=\log (x+y)$, then $\frac{d y}{d x}=$

A. 2

B. $-2$

C. 1

D. $-1$

Solution:

$\sin (x+y)=\log (x+y)$

Differentiating w.r.t $\mathrm{x}$ we get,

$\Rightarrow \cos (x+y) \cdot\left(1+\frac{d y}{d x}\right)=\frac{1}{x+y} \cdot\left(1+\frac{d y}{d x}\right)$

$\Rightarrow \cos (x+y) \cdot\left(1+\frac{d y}{d x}\right)-\frac{1}{x+y} \cdot\left(1+\frac{d y}{d x}\right)=0$

$\Rightarrow\left(1+\frac{\mathrm{dy}}{\mathrm{dx}}\right)\left(\cos (\mathrm{x}+\mathrm{y})-\frac{1}{\mathrm{x}+\mathrm{y}}\right)=0$

$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\left(\cos (\mathrm{x}+\mathrm{y})-\frac{1}{\mathrm{x}+\mathrm{y}}\right)+\left(\cos (\mathrm{x}+\mathrm{y})-\frac{1}{\mathrm{x}+\mathrm{y}}\right)=0$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\left(\cos (\mathrm{x}+\mathrm{y})-\frac{1}{\mathrm{x}+\mathrm{y}}\right)}{\left(\cos (\mathrm{x}+\mathrm{y})-\frac{1}{\mathrm{x}+\mathrm{y}}\right)}=-1$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-1$