Question

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

• A. $\tan \left(x+y\right)$
• B. ${\log }_e\left(x+y\right)$
• C. $-1$
• D. $1$

Answer 1

The correct option is C

$-1$

Explanation for the correct option

Step 1: Differentiate with respect to $x$

Given information

$\therefore \sin (x+y)={\log }_e(x+y)$

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

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

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

Step 2: Take common $\left(1+\frac{dy}{dx}\right)$ and solve the above equation

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

$\Rightarrow \left(1+\frac{dy}{dx}\right)=0$

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

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

Hence option (C) is the correct answer.