Question

Differentiate the function ${\sin }^{3}x+{\cos }^{6}x$ w.r.t. $x$

Answer 1

Let $y={\sin }^{3}x+{\cos }^{6}x$

Differentiating w.r.t. $x$, we get

$\frac{dy}{dx}=\frac{d({\sin }^{3}x+{\cos }^{6}x)}{dx}$

$\frac{dy}{dx}=\frac{d\left(\right({\sin }^{3}x\left)\right)}{dx}+\frac{d\left(\right({\cos }^{6}x\left)\right)}{dx}$

$\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$

$\frac{dy}{dx}=3{\sin }^{2}x.\frac{d\left(\sin x\right)}{dx}+6{\cos }^{5}x.\frac{d\left(\cos x\right)}{dx}$

$\frac{dy}{dx}=3{\sin }^{2}x.\cos x+6{\cos }^{5}x.(-\sin x)$

$\frac{dy}{dx}=3{\sin }^{2}x.\cos x-6{\cos }^{5}x.\sin x$