Question

(cosx)^y = (siny)^x find dy/dx​

Answer 1

Step-by-step explanation:

We Have :-

$\cos(x)^{y} = \sin(y)^{x}$

To Find :-

$\dfrac{dy}{dx}$

Method Used :-

$taking \: log \: both \: sides$

Solution:-

$\cos(x)^{y} = \sin(y)^{x} \\ \\ \\ taking \: log \\ \\ \\ log( \cos(x)^{y} ) = log( \sin(y)^{x} ) \\ \\ \\ y log( \cos(x) ) = x log( \sin(y) ) \\ \\ \\ using \: multiplication \: rule \: of \: differentiation \\ \\ \\ - y \dfrac{ \sin(x) }{ \cos(x) } + log( \cos(x) ) \dfrac{dy}{dx} = log( \sin(y) ) + y \dfrac{ \cos(y) }{ \sin(y) } \dfrac{dy}{dx} \\ \\ \\ y \tan(x) + log( \sin(y) ) = \dfrac{dy}{dx} (y \cot(y) - log( \cos(x) ) ) \\ \\ \\ \dfrac{dy}{dx} = \dfrac{y \tan(x) + log( \sin(y) ) }{y \cot(y) - log( \cos(x) ) }$