Question

y= log sinx + log cosx find dy/dx​

Answer 1

Answer :

y = log sinx + log cosx

dy/dx = cosx/sinx - sinx/cosx

therefore :

dy/dx = (cos^2x - sin^2x) / sinx cosx

multiply numerator and denominator with 2

and cos^2x - sin^2x = cos2x

therefore :

dy/dx = 2cos2x/2sinxcosx

dy/dx = 2cos2x/sin2x

dy/dx = 2tan2x

Answer 2

Answer:

$\frac{dy}{dx}$ = cotx - tanx

Step-by-step explanation:

$y=log(sinx)+log(cosx)\\\\\frac{dy}{dx} =\frac{1}{sinx}.(cosx)(1) + \frac{1}{cosx} .(-sinx)(1)\\\\\frac{dy}{dx}= \frac{cosx}{sinx} -\frac{sinx}{cosx} \\\\\frac{dy}{dx}=cotx-tanx$