Answer
$\frac{dy}{dx} = -\frac{1}{x} sin ^{-5}(\frac{1}{x} + 5sin ^{-1}x cos x) - cos^{2}x (\frac{1}{3}cos x -x sin x)$
Work Step by Step
$y = \frac{1}{x}u^{-5} - \frac{x}{3}v^{3}$
where,
$u = sin x$
$v = cos x$
and,
$u' = cos x$
$v' = -sin x$
Now,
$y = a-b$,
where,
$a = \frac{1}{x}u^{-5}$
$b = \frac{x}{3}v^{3}$
Differentiate these two terms:
$a' = \frac{-1}{x^{2}}u^{-5} -5 \frac{1}{x}u^{-6} \times (cos x)$
$ = \frac{-1}{x^{2}}sin^{-5} x -5 \frac{1}{x}sin^{-6} x cos x$
$ = -\frac{1}{x} sin ^{-5}(\frac{1}{x} + 5sin ^{-1}x cos x)$
$b' = \frac{1}{3}v^{3} + xv^{2} \times (-sin x) = \frac{1}{3}cos^{3}x - xcos^{2}x sin x$
$b' = cos^{2}x (\frac{1}{3}cos x -x sin x)$
So now,
$\frac{dy}{dx} = a' - b'$
$\frac{dy}{dx} = -\frac{1}{x} sin ^{-5}(\frac{1}{x} + 5sin ^{-1}x cos x) - cos^{2}x (\frac{1}{3}cos x -x sin x)$
