find the derivative of the function.\n\n$y = \\sin^{2}(x^{2}+7)$

Explanation:

Step1: Let (u = x^{2}+7) and (y=\sin^{2}u)

We will use the chain - rule (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}).

Step2: First, find (\frac{dy}{du}) for (y = \sin^{2}u)

Let (t=\sin u), then (y = t^{2}). By the chain - rule, (\frac{dy}{du}=\frac{dy}{dt}\cdot\frac{dt}{du}). For (y = t^{2}), (\frac{dy}{dt}=2t). For (t=\sin u), (\frac{dt}{du}=\cos u). So (\frac{dy}{du}=2\sin u\cos u=\sin(2u)).

Step3: Then, find (\frac{du}{dx}) for (u=x^{2}+7)

Using the power - rule, (\frac{du}{dx}=2x).

Step4: Now, find (\frac{dy}{dx}) using the chain - rule (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx})

Substitute (u = x^{2}+7), (\frac{dy}{du}=\sin(2u)=\sin(2(x^{2}+7))) and (\frac{du}{dx}=2x) into the chain - rule formula. So (\frac{dy}{dx}=2x\sin(2(x^{2}+7))).

Answer:

(2x\sin(2(x^{2}+7)))