Question

find the derivative of the given function.
y = cos (e^{-4θ^{3}})
\frac{dy}{dθ} =

Explanation:

Step1: Let \(u = e^{-4\theta^{3}}\)

The function \(y=\cos(u)\). By the chain - rule \(\frac{dy}{d\theta}=\frac{dy}{du}\cdot\frac{du}{d\theta}\).
First, find \(\frac{dy}{du}\). Since \(y = \cos(u)\), then \(\frac{dy}{du}=-\sin(u)\).

Step2: Find \(\frac{du}{d\theta}\)

Since \(u = e^{-4\theta^{3}}\), by the chain - rule again. Let \(v=-4\theta^{3}\), then \(u = e^{v}\). \(\frac{du}{dv}=e^{v}\) and \(\frac{dv}{d\theta}=-12\theta^{2}\). So \(\frac{du}{d\theta}=\frac{du}{dv}\cdot\frac{dv}{d\theta}=e^{v}\cdot(- 12\theta^{2})\). Substituting \(v = - 4\theta^{3}\), we get \(\frac{du}{d\theta}=-12\theta^{2}e^{-4\theta^{3}}\).

Step3: Calculate \(\frac{dy}{d\theta}\)

Substitute \(\frac{dy}{du}\) and \(\frac{du}{d\theta}\) into the chain - rule formula \(\frac{dy}{d\theta}=\frac{dy}{du}\cdot\frac{du}{d\theta}\).
\(\frac{dy}{d\theta}=-\sin(e^{-4\theta^{3}})\cdot(-12\theta^{2}e^{-4\theta^{3}})\)

Answer:

\(12\theta^{2}e^{-4\theta^{3}}\sin(e^{-4\theta^{3}})\)