Answer
$$y'= 4{[x+{(x+\sin^2 x)}^3]}^3
[3{(x+\sin^2 x)}^2
(1+2\sin x\cos x)+1]$$
Work Step by Step
$y' = \frac{d}{dx} {[x+{(x+\sin^2 x)}^3]}^4$
By the chain rule:
$= \frac{d{[x+{(x+\sin^2 x)}^3]}^4}{d[x+{(x+\sin^2 x)}^3]}
\times \frac{d[x+{(x+\sin^2 x)}^3]}{dx}$
$= 4{[x+{(x+\sin^2 x)}^3]}^3
\times [\frac{d}{dx} [x]+\frac{d}{dx}[{(x+\sin^2 x)}^3]]$
Using the chain rule again:
$= 4{[x+{(x+\sin^2 x)}^3]}^3
\times [1+\frac{d[{(x+\sin^2 x)}^3]}{d[{(x+\sin^2 x)}]}
\times\frac{d[{(x+\sin^2 x)}]}{dx}]$
$= 4{[x+{(x+\sin^2 x)}^3]}^3
\times [1+3{(x+\sin^2 x)}^2
(1+2\sin x\cos x)]$
$= 4{[x+{(x+\sin^2 x)}^3]}^3
[3{(x+\sin^2 x)}^2
(1+2\sin x\cos x)+1]$
