Question

find the derivative of: $-6\sin^{2}(3x^{3})$.
hint: $\sin^{2}(x)=\sin(x)^{2}$...so use the chain rule (twice!)

Explanation:

Step1: Apply the chain rule for the outer function

Let \(u = \sin(3x^{3})\), then the function is \(y=-6u^{2}\). The derivative of \(y\) with respect to \(u\) is \(\frac{dy}{du}=-12u\).

Step2: Apply the chain rule for the inner - function

Now, find the derivative of \(u = \sin(3x^{3})\) with respect to \(x\). Let \(v = 3x^{3}\), 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 \(\frac{dv}{dx}=9x^{2}\). By the chain rule \(\frac{du}{dx}=\frac{du}{dv}\cdot\frac{dv}{dx}=\cos(3x^{3})\cdot9x^{2}\).

Step3: Combine the results using the chain rule

By the chain rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\). Substitute \(\frac{dy}{du}=-12\sin(3x^{3})\) and \(\frac{du}{dx}=9x^{2}\cos(3x^{3})\) into the formula.

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Answer:

\(-108x^{2}\sin(3x^{3})\cos(3x^{3})\)