find the derivative of the function.\n$y = \\sin^{2}(x^{2}+8)$\n$y=$

Explanation:

Step1: Use chain - rule for outer function

Let $u=\sin(x^{2}+8)$. Then $y = u^{2}$. The derivative of $y$ with respect to $u$ is $\frac{dy}{du}=2u$.

Step2: Use chain - rule for inner function

Now find the derivative of $u=\sin(x^{2}+8)$ with respect to $x$. Let $v=x^{2}+8$. Then $u = \sin(v)$. The derivative of $u$ with respect to $v$ is $\frac{du}{dv}=\cos(v)$, and the derivative of $v$ with respect to $x$ is $\frac{dv}{dx}=2x$.

Step3: Apply the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dv}\cdot\frac{dv}{dx}$

Substitute back $u=\sin(x^{2}+8)$ and $v=x^{2}+8$ into the chain - rule formula. $\frac{dy}{dx}=2u\cdot\cos(v)\cdot2x=2\sin(x^{2}+8)\cdot\cos(x^{2}+8)\cdot2x$. Using the double - angle formula $\sin(2\alpha)=2\sin\alpha\cos\alpha$, we can simplify it to $y' = 4x\sin(x^{2}+8)\cos(x^{2}+8)=2x\sin(2(x^{2}+8))$.

Answer:

$2x\sin(2(x^{2}+8))$