Question

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

Answer 1

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

$\frac{d}{dx}\left[\sin \right(x+y\left)\right]=\frac{d}{dx}\left[\log \right(x+y\left)\right]$

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

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

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

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

$\Rightarrow$ $1+\frac{dy}{dx}=0$

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