Answer
$\frac{dy}{dx}$ = -$\frac{5}{x}$$sin^{-6}x$($cosx$) - ($\frac{1}{x^{2}}$)$sin^{-5}x$ + $x$$cos^{2}x$($sinx$) - $\frac{1}{3}$$cox^{3}x$
Work Step by Step
$y$ = $\frac{1}{x}$$sin^{-5}(x)$ - $\frac{x}{3}$$cos^{3}(x)$
$y$ = $\frac{1}{x}$$(sinx)^{5}$ - $\frac{x}{3}$$(cosx)^{3}$
differentiate with respect to x
$\frac{dy}{dx}$ = $\frac{d}{dx}$[$\frac{1}{x}$$(sinx)^{-5}$] - $\frac{d}{dx}$[$\frac{x}{3}$$(cosx)^{3}$]
use the product rule for each term
$\frac{dy}{dx}$ = $\frac{1}{x}$$\frac{d}{dx}$[$(sinx)^{-5}$] + $(sinx)^{-5}$$\frac{d}{dx}$[$\frac{1}{x}$] - $\frac{x}{3}$$\frac{d}{dx}$[$(cosx)^{3}$] - $(cosx)^{3}$$\frac{d}{dx}$[$\frac{x}{3}$]
solve derivative using the chain rule for
$\frac{dy}{dx}$ = $\frac{1}{x}$$(-5)(sinx)^{-6}$$\frac{d}{dx}$($sinx$) + $(sinx)^{-5}$($-\frac{1}{x^{2}}$) - $\frac{x}{3}$$(3)(cosx)^{2}$$\frac{d}{dx}$[$(cosx)$] - $(cosx)^{3}$$\frac{1}{3}$
$\frac{dy}{dx}$ = $\frac{-5}{x}$$(sinx)^{-6}$($cosx$) - $(sinx)^{-5}$($\frac{1}{x^{2}}$) - $x$$(cosx)^{2}$(-$sinx$) - $(cosx)^{3}$$\frac{1}{3}$
$\frac{dy}{dx}$ = -$\frac{5}{x}$$sin^{-6}x$($cosx$) - ($\frac{1}{x^{2}}$)$sin^{-5}x$ + $x$$cos^{2}x$($sinx$) - $\frac{1}{3}$$cox^{3}x$
simplify
$\frac{dy}{dx}$ = -$\frac{5}{x}$$sin^{-6}x$($cosx$) - ($\frac{1}{x^{2}}$)$sin^{-5}x$ + $x$$cos^{2}x$($sinx$) - $\frac{1}{3}$$cox^{3}x$
