Question

If y = (cos x) ^logx,​ find dy/dx

Answer 1

Step-by-step explanation:

We have,

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

Taking log both sides,

$log(y) = log(x) . log( \cos(x) )$

Differentiating both sides w.r.t x, we have,

$\implies \frac{1}{y} . \frac{dy}{dx} = log( \cos(x) ). \frac{d}{dx}( log(x) ) + log(x). \frac{d}{dx} ( log( \cos(x) ) )$

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

$\implies \frac{dy}{dx} = y. [ \frac{ log( \cos(x) ) }{x} - \tan(x) log(x) ]$

$\implies \frac{dy}{dx} = (\cos(x) )^{ log(x) } . [ \frac{ log( \cos(x) ) }{x} - \tan(x) log(x) ]$