Question

Find $\frac{dy}{dx}$, if $x+y=\sin (x-y)$

• A. $\frac{\cos (x-y)-1}{\cos (x-y)+1}$
• B. $\frac{\cos (x-y)+1}{\cos (x-y)-1}$
• C. $\frac{\cos (x+y)+1}{\cos (x-y)-1}$
• D. $\frac{\cos (x+y)-1}{\cos (x-y)+1}$

Answer 1

The correct option is A $\frac{\cos (x-y)-1}{\cos (x-y)+1}$

Given, $x+y=\sin (x-y)$

Differentiate both side w.r.t $x$

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

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