Finding a derivative using multiple chain rules

Question

Find the derivative of the function. $y = [x + (x + \sin^2 x)^3]^4$

I know how to use the chain rule and I found the derivative to be:

$$4[x+(x+\sin^2(x))^3]^3 \cdot (1 + 3(x + \sin^2(x))^2) \cdot (1+\sin (2x))$$

but my online homework says that this is wrong. I can't figure what what I've done wrong and I've tried it several times now. Can somebody help?

Note: In the last term, I simplified $2\sin x\cos x$ to be $\sin(2x)$. I tried inputting both versions into my homework but it was wrong both ways.

Answer 1

\begin{align} \textbf{wrong:} \quad & 4[x+(x+\sin^2(x))^3]^3 \cdot \Big(1 + 3(x + \sin^2(x))^2\Big) \cdot (1+\sin (2x)) \\[10pt] \textbf{right:} \quad & 4[x+(x+\sin^2(x))^3]^3 \cdot \Big(1 + 3(x + \sin^2(x))^2 \cdot(1+\sin (2x)) \Big) \end{align} The thing that gets multiplied by $1+\sin(2x)$ should be only $3(x+\sin^2 x)^2,$ not the sum of that and $1$.

Answer 2

It might be a good idea to introduce some new functions to avoid messing up with the calculations. Consider $$ y=g^4(x) $$ where $$ g(x)=x+h^3(x) \quad\textrm{and }\ h(x)=x + \sin^2 x. $$ Then $$ y'=4g^3(x)g'(x) $$ where $$ g'(x)=1+3h^2(x)\cdot h'(x) $$ and $$ h'(x)=1+2\sin x\cos x. $$ Putting everything together you have the answer: $$ y'=4g^3(x)\bigr(1+3h^2(x)\cdot (1+2\sin x\cos x)\bigr) $$