Question

If $Sin(x+y)=\log (x+y)$ then $\frac{dy}{dx}$is equal to

• A. $2$
• B. $-2$
• C. $1$
• D. $-1$

Answer 1

The correct option is D

$-1$

Explanation for the correct option

Step 1:Differentiate with respect to $x$

Given information

$\Rightarrow Sin(x+y)=\log (x+y)$

$\therefore \cos (x+y)\left(1+\frac{dy}{dx}\right)=\left[\frac{1}{\left(x+y\right)}\right]\left(1+\frac{dy}{dx}\right)$

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

Step 2: Compare on both sides and take differential at one side

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

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

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

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

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

Therefore the value of $\frac{dy}{dx}$ is $-1$

Hence option (4) is the correct answer.