Problem 51 - Calculus - AP Edition

Question

Find \(d^{2} y / d x^{2}\) $$y=x \cos (5 x)-\sin ^{2} x$$

Step-by-Step Solution

Answer

The second derivative is \(-10\sin(5x) - 25x\cos(5x) + 2\cos(2x)\)."

Step 1: Identify Components for Differentiation

The function given is \(y = x \cos(5x) - \sin^2(x)\). We need to take the second derivative of this function with respect to \(x\), which requires finding the first derivative \(dy/dx\) first. Notice that the expression has two main components: \(x \cos(5x)\) and \(-\sin^2(x)\).

Step 2: Differentiate the first component

Start by differentiating \(x \cos(5x)\). Use the product rule for differentiation, which states \(d(uv)/dx = u'v + uv'\). Let \(u = x\) and \(v = \cos(5x)\), then \(u' = 1\) and \(v' = -5\sin(5x)\). So, \(d(x \cos(5x))/dx = 1 \cdot \cos(5x) + x(-5\sin(5x)) = \cos(5x) - 5x\sin(5x)\).

Step 3: Differentiate the second component

Now differentiate \(-\sin^2(x)\). Use the chain rule: if \(f(x) = -\sin^2(x)\), then let \(u = \sin(x)\) so \(f = -u^2\). The derivative is \(df/dx = d(-u^2)/du \, \cdot \, du/dx = -2u \cdot \cos(x) = -2\sin(x)\cos(x)\).

Step 4: Combine first derivatives

Combine the results from Steps 2 and 3 to find the first derivative: \(dy/dx = \cos(5x) - 5x\sin(5x) - 2\sin(x)\cos(x)\).

Step 5: Differentiate again for the second derivative

Differentiate the expression from Step 4 to find \(d^2y/dx^2\). Use the sum rule and product rule where necessary. Differentiate each term separately: 1. \(d/dx[\cos(5x)] = -5\sin(5x)\).2. \(d/dx[-5x\sin(5x)] = -5[1 \cdot \sin(5x) + x \cdot 5\cos(5x)] = -5\sin(5x) - 25x\cos(5x)\).3. \(d/dx[-2\sin(x)\cos(x)] = -2[-\sin^2(x) + \cos^2(x)] = 2(\cos^2(x) - \sin^2(x)) = 2\cos(2x)\) using the double-angle identity.

Step 6: Combine second derivative components

Combine the results from Step 5 to get the second derivative: \(\frac{d^2y}{dx^2} = -5\sin(5x) - 5\sin(5x) - 25x\cos(5x) + 2\cos(2x) = -10\sin(5x) - 25x\cos(5x) + 2\cos(2x)\).

Key Concepts

Product Rule

The product rule is a cornerstone of calculus used when differentiating products of two functions. When given a function of the form \(y = u(x)v(x)\), the derivative \(dy/dx\) is found using the formula

• \(\frac{d}{dx}[uv] = u'v + uv'\).

In the context of the original exercise, the term \(x\cos(5x)\) required us to apply the product rule. Here, we defined \(u = x\) and \(v = \cos(5x)\). Consequently, the derivatives were \(u' = 1\) and \(v' = -5\sin(5x)\).
This led to:

• \(\frac{d}{dx}[x\cos(5x)] = 1 \cdot \cos(5x) + x (-5\sin(5x)) = \cos(5x) - 5x\sin(5x)\).

This step shows how combining the derivatives of \(u\) and \(v\) correctly contributes to finding \(dy/dx\) accurately.

Chain Rule

The chain rule is instrumental when differentiating compositions of functions. It explains how to find the derivative of a function \(f(g(x))\) as

• \(f'(g(x)) \cdot g'(x)\).

In our exercise, we needed the chain rule for \(-\sin^2(x)\). Letting \(u = \sin(x)\) transformed the expression into \(-u^2\). This required

• \(\frac{d}{dx}[-\sin^2(x)] = -2u \cdot \frac{du}{dx} = -2\sin(x) \cdot \cos(x)\).

The chain rule simplifies differentiations where basic algebraic manipulations and direct differentiation rules cannot be applied straightforwardly. This helps achieve the correct first derivative component for \(-\sin^2(x)\).

Trigonometric Functions

Trigonometric functions like \(\sin(x)\) and \(\cos(x)\) are common in calculus, and understanding their derivatives is vital. For instance, the derivatives are:

• \(\frac{d}{dx}[\sin(x)] = \cos(x)\)
• \(\frac{d}{dx}[\cos(x)] = -\sin(x)\)

In our solution, these properties helped differentiate terms like \(\cos(5x)\) and those involving products like \(-\sin^2(x)\). The term \(\cos(2x)\) derived during the second differentiation results from trigonometric identities like the double-angle formula:

• \(\cos(2x) = \cos^2(x) - \sin^2(x)\).

Such identities are useful in simplifying and accurately transforming expressions that's needed to complete tasks like finding \(d^2y/dx^2\) in a manageable way.