Question

Differentiate w.r.t. $x$ the function: $sin^3x+cos^6x$

Answer 1

The correct answer is $=3sinxcosx(sinx-2cos^4x)$
Let $y=sin^3x+cos^6x$
$∴\frac{dy}{dx}=\frac{d}{dx}(sin^3x)+\frac{d}{dx}(cos^6x)$
$=3sin^2x.\frac{d}{dx}(sinx)+6cos^5x.\frac{d}{dx}(cosx)$
$=3sin^2x.cosx+6cos^5x.(-sinx)$
$=3sinxcosx(sinx-2cos^4x)$