Question

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

Answer 1

$x+y=sin(x-y)=sinxcosy-cosxsiny$

Differentiating with respect to x,

$1+\frac{dy}{dx}=sinx(-siny)\frac{dy}{dx}+cosxcosy$

$-cosx\left(cosy\right)\frac{dy}{dx}+sinxsiny$

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

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

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