Problem 35 - Calculus Single Variable

Question

Calculate the requested derivative. \(g^{\prime \prime}(\pi / 6)\) where \(g(x)=\cos (\pi \sin (x))\)

Step-by-Step Solution

Answer

The second derivative, \(g''(\pi/6)\), is \(3\pi^2/4 + \pi/2\).

Step 1 - Differentiate g(x) once

We start with the function \(g(x) = \cos(\pi \sin(x))\). First, we need to find the first derivative, \(g'(x)\), using the chain rule. The derivative of \(\cos(u)\) with respect to \(u\) is \(-\sin(u)\), and the derivative of \(\pi \sin(x)\) with respect to \(x\) is \(\pi \cos(x)\). Therefore, \(g'(x) = -\sin(\pi \sin(x)) \cdot \pi \cos(x)\).

Step 2 - Differentiate g'(x) to find g''(x)

Differentiate \(g'(x) = -\pi \sin(\pi \sin(x)) \cdot \cos(x)\) to find \(g''(x)\). Use the product rule, which is \((u v)' = u' v + u v'\). Let \(u = -\pi \sin(\pi \sin(x))\) and \(v = \cos(x)\). So, \(u' = -\pi^2 \cos(\pi \sin(x)) \cdot \cos(x)\) and \(v' = -\sin(x)\). Thus, \(g''(x) = -\pi^2 \cos(\pi \sin(x)) \cos^2(x) + \pi \sin(\pi \sin(x)) \sin(x)\).

Step 3 - Evaluate g''(x) at x = π/6

Now, substitute \(x = \pi / 6\) into \(g''(x)\). First, calculate \(\sin(\pi / 6) = 1/2\) and \(\cos(\pi / 6) = \sqrt{3}/2\). Substitute these into \(g''(x)\): \(-\pi^2 \cos(\pi \cdot 1/2) \cdot \left(\sqrt{3}/2\right)^2 + \pi \sin(\pi \cdot 1/2) \cdot 1/2\). This simplifies to \(-\pi^2 \cdot (-1) \cdot 3/4 + \pi \cdot 1 \cdot 1/2\), which evaluates to \(3\pi^2/4 + \pi/2\).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in derivative calculus used to calculate the derivative of composite functions. It allows us to differentiate a function based on its constituent parts. Imagine you have a function inside another function, like nesting dolls.
The beauty of the chain rule is that it helps simplify the differentiation of such complex functions by breaking them down.

• Consider a function \( h(x) = f(g(x)) \), where \( f \) is an outer function and \( g \) is an inner function.
• The chain rule states: \( h'(x) = f'(g(x)) \cdot g'(x) \).

In our exercise, we used the chain rule to differentiate \( g(x) = \cos(\pi \sin(x)) \). Here, \( u = \pi \sin(x) \) acts as the inner function, and \( \cos(u) \) is the outer function.
First, we take the derivative of \( \cos(u) \), which is \(-\sin(u)\), and multiply it by the derivative of the inner function \( \pi \sin(x) \), which results in \( \pi \cos(x) \). Thus, yielding the derivative as \( g'(x) = -\sin(\pi \sin(x)) \cdot \pi \cos(x) \).

Product Rule

The product rule is a method for differentiating products of two or more functions. Remember, when you multiply two functions, each carries their own rate of change.
The product rule gives us the formula to find the derivative of such functions efficiently.

• If you have functions \( u(x) \) and \( v(x) \), the product rule states: \((uv)' = u'v + uv'\).
• It's crucial when you're working with expressions involving multiplication of different derivatives.

In our context, after we found \( g'(x) \), we needed to apply the product rule to find \( g''(x) \) as it involves a product of functions.
Letting \( u = -\pi \sin(\pi \sin(x)) \) and \( v = \cos(x) \), we computed their derivatives as \( u' = -\pi^2 \cos(\pi \sin(x)) \cos(x) \) and \( v' = -\sin(x) \),
then substituted these values into the formula: \( g''(x) = -\pi^2 \cos(\pi \sin(x)) \cos^2(x) + \pi \sin(\pi \sin(x)) \sin(x) \).

Trigonometric Derivatives

Trigonometric derivatives involve taking the derivative of trigonometric functions like sine, cosine, and tangent. These derivatives are foundational as they frequently appear in calculus problems and real-world applications involving waves, circles, and oscillations.
Key trigonometric derivatives to remember:

• Derivative of \( \sin(x) \) is \( \cos(x) \)
• Derivative of \( \cos(x) \) is \( -\sin(x) \)
• Derivative of \( \tan(x) \) is \( \sec^2(x) \)

In our original exercise, the knowledge of trigonometric derivatives was imperative.
We differentiated \( \cos(\pi \sin(x)) \) first using \( \cos(u) \)'s derivative, \( -\sin(u) \), helping us find \( g'(x) = -\sin(\pi \sin(x)) \cdot \pi \cos(x) \).
These trigonometric derivatives are essential tools, helping transform complex periodic functions into simpler linear relationships, which are much easier to analyze and solve.