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

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.

[ \begin{align*} \frac{dy}{dx}&=-12\sin(3x^{3})\cdot9x^{2}\cos(3x^{3})\ &=- 108x^{2}\sin(3x^{3})\cos(3x^{3}) \end{align*} ]

Answer:

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