Question

Find $\frac{d}{dx}{\sin }^{3}4x{\cos }^{8}5x$

Answer 1

$\frac{d}{dx}\left({\sin }^{3}4x{\cos }^{8}5x\right)$

Let $u={\sin }^{3}4x$ and $v={\cos }^{8}5x$

$\therefore \frac{d}{dx}\left({\sin }^{3}4x{\cos }^{8}5x\right)=\frac{d}{dx}\left(uv\right)=u\frac{dv}{dx}+v\frac{du}{dx}$

$={\sin }^{3}4x\cdot \frac{d}{dx}\left({\cos }^{8}5x\right)+{\cos }^{8}5x\frac{d}{dx}\left({\sin }^{3}4x\right)$ …..$\left(1\right)$

Here $\frac{d}{dx}\left({\cos }^{8}5x\right)=\frac{d}{d\left(\cos 5x\right)}\left\{\right(\cos 5x{)}^8\}\cdot \frac{d\left(\cos 5x\right)}{dx}$

$=8(\cos 5x{)}^7\cdot \frac{d}{d\left(5x\right)}(\cos \left(5x\right))=\frac{d\left(5x\right)}{dx}$

$=8{\cos }^{-1}5x\cdot (-\sin 5x)\cdot 5\frac{dx}{dx}$ $[\because \frac{d\left(\cos x\right)}{dx}=-\sin x]$

$=-40\sin 5x{\cos }^{7}5x$.

and $\frac{d}{dx}\left({\sin }^{3}4x\right)=\frac{d}{d\left(\sin 4x\right)}\left\{\right(\sin 4x{)}^3\}.\frac{d\left(\sin 4x\right)}{dx}$

$=3{\sin }^{2}4x\cdot \frac{d\left(\sin 4x\right)}{d\left(4x\right)}\cdot \frac{d\left(4x\right)}{dx}$

$=3{\sin }^{2}4x\cos 4x4\frac{dx}{dx}$ $[\because \frac{d}{dx}(\sin x)=\cos x]$

$=12{\sin }^{2}4x\cos 4x$

From $\left(1\right)$,

$\frac{d}{dx}{\sin }^{3}4x{\cos }^{8}5x={\sin }^{3}4x.(-40\sin 5x{\cos }^{7}5x)+{\cos }^{8}5x\left(12{\sin }^{2}4x\cos 4x\right)$

$=-40{\sin }^{3}4x\sin 5x{\cos }^{7}5x+12{\cos }^{8}5x{\sin }^{2}4x\cos 4x$

$=4(-10{\sin }^{3}4x\sin 5x{\cos }^{7}5x+3{\sin }^{2}4x\cos 4x{\cos }^{8}5x)$

$=4{\cos }^{7}5x\cdot {\sin }^{2}4x(3\cos 4x\cos 5x-10\sin 4x\sin 5x)$

$=4{\cos }^{7}5x{\sin }^{2}4x(\frac{3\cos (4x+5x)+\cos (4x-5x)}{2}-10\times \frac{\cos (4x-5x)-\cos (4x+5x)}{2})$

$=4{\cos }^{7}5x{\sin }^{2}4x\left(\frac{3\cos 9x+3\cos (-x)-10\cos (-x)+\cos 9x}{2}\right)$

$=2{\cos }^{7}5x{\sin }^{2}4x(4\cos 9x-7\cos x)$ $[\because \cos (-x)=\cos x]$

$\therefore \frac{d}{dx}\left({\sin }^{3}4x{\cos }^{8}5x\right)=2{\cos }^{7}5x{\sin }^{2}4x(4\cos 9x-7\cos x)$.