Question Statement

Find the third derivative of the function $f(x)=\cos (\pi x)$, denoted as $f^{\prime \prime \prime }(x)$.

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Background and Explanation

This problem requires applying the chain rule repeatedly to find higher-order derivatives of a trigonometric function with a linear argument. Recall that the derivatives of cosine and sine follow cyclic patterns, and each differentiation introduces an additional factor of $\pi$ from the inner function.

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Solution

We begin with the given function: $f(x)=\cos (\pi x)$

First Derivative

Differentiate with respect to $x$ using the chain rule. Let $u=\pi x$, so $\frac{du}{dx}=\pi$:

$\begin{array}{rl}{f}^{\prime }(x)& =\frac{d}{dx}[\cos (\pi x)]\\ & =-\sin (\pi x)\cdot \frac{d}{dx}(\pi x)\\ & =-\sin (\pi x)\cdot \pi \\ & =-\pi \sin (\pi x)\end{array}$

Second Derivative

Differentiate $f^{\prime }(x)=-\pi \sin (\pi x)$ with respect to $x$, again applying the chain rule:

$\begin{array}{rl}{f}^{\prime \prime }(x)& =\frac{d}{dx}[-\pi \sin (\pi x)]\\ & =-\pi \cdot \cos (\pi x)\cdot \frac{d}{dx}(\pi x)\\ & =-\pi \cdot \cos (\pi x)\cdot \pi \\ & =-{\pi }^2\cos (\pi x)\end{array}$

Third Derivative

Differentiate $f^{\prime \prime }(x)=-{\pi }^2\cos (\pi x)$ with respect to $x$:

$\begin{array}{rl}{f}^{\prime \prime \prime }(x)& =\frac{d}{dx}[-{\pi }^2\cos (\pi x)]\\ & =-{\pi }^2\cdot (-\sin (\pi x))\cdot \frac{d}{dx}(\pi x)\\ & ={\pi }^2\sin (\pi x)\cdot \pi \\ & ={\pi }^3\sin (\pi x)\end{array}$

Therefore, the third derivative is: $f^{\prime \prime \prime }(x)={\pi }^3\sin (\pi x)$

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Key Formulas or Methods Used

• Chain Rule: $\frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))\cdot g^{\prime }(x)$
• Derivative of Cosine: $\frac{d}{dx}[\cos (x)]=-\sin (x)$
• Derivative of Sine: $\frac{d}{dx}[\sin (x)]=\cos (x)$
• Power Rule for Constants: $\frac{d}{dx}[\pi x]=\pi$

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Summary of Steps

1. First derivative: Apply chain rule to get $f^{\prime }(x)=-\pi \sin (\pi x)$
2. Second derivative: Differentiate again to obtain $f^{\prime \prime }(x)=-{\pi }^2\cos (\pi x)$
3. Third derivative: Differentiate once more to arrive at $f^{\prime \prime \prime }(x)={\pi }^3\sin (\pi x)$