Question

Implicite function
If sin y = log(x+y) then dy/dx.

Answer 1

Answer:

$\bold{ \dfrac{dy}{dx} = \dfrac{1}{(x + y) \: cosy \: - 1} }$

Step-by-step explanation:

$\bold{To \: find} = > derivative \: of \: \\ siny = log(x + y)$

$\bold{Concept \: used }= > \\ 1) \: \dfrac{d}{dx} (sinx) =cosx \\ 2) \: \dfrac{d}{dx} ( logx) = \dfrac{1}{x}$

$\bold{Solution} = > \\ siny = log(x + y)$

$differentiating \: with \: respect \: to \: x$

$= > \dfrac{d}{dx} (siny) = \dfrac{d}{dx} log(x + y)$

$= > cosy \: \dfrac{dy}{dx} = \dfrac{1}{x + y} \: \dfrac{d}{dx} (x + y)$

$= > (x + y) \: cosy \: \dfrac{dy}{dx} = \dfrac{d}{dx} (x) + \dfrac{dy}{dx}$

$= > (x + y) \: cosy \: \dfrac{dy}{dx} - \dfrac{dy}{dx} = 1$

$= > ( \: (x + y)cosy \: - 1 \: ) \: \dfrac{dy}{dx} \: = 1$

$= > \: \dfrac{dy}{dx} = \dfrac{1}{(x + y) \: cosy \: - 1}$