Question

1. $y = cos^{3}(pi x)$

Explanation:

Step1: Identify the outer and inner functions

Let \(u = \cos(\pi x)\), so \(y = u^{3}\).

Step2: Differentiate the outer function

Using the power rule \((x^{n})^\prime=nx^{n - 1}\), \(\frac{dy}{du}=3u^{2}\).

Step3: Differentiate the inner function

Let \(v=\pi x\), then \(u = \cos(v)\). First, \(\frac{du}{dv}=-\sin(v)\), and \(\frac{dv}{dx}=\pi\). By the chain - rule \(\frac{du}{dx}=\frac{du}{dv}\cdot\frac{dv}{dx}\), so \(\frac{du}{dx}=-\pi\sin(\pi x)\).

Step4: Apply the chain rule

By the chain rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\). Substitute \(\frac{dy}{du}=3u^{2}\) and \(u = \cos(\pi x)\), \(\frac{du}{dx}=-\pi\sin(\pi x)\) into it.
\(\frac{dy}{dx}=3\cos^{2}(\pi x)\cdot(-\pi\sin(\pi x))\)

Answer:

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