Problem 22 - Calculus - AP Edition

Question

Find \(d^{2} y / d x^{2}\) $$y=x^{2} \cos x+4 \sin x$$

Step-by-Step Solution

Answer

The second derivative is \(-4x \sin x - 2 \cos x - x^2 \cos x - 4 \sin x\)."

Step 1: Differentiate the First Term

Start by differentiating the first term of the function, which is \(x^2 \cos x\). We need to use the product rule, which states \(d(uv)/dx = u'v + uv'\). Let \(u=x^2\) and \(v=\cos x\). So, \(u' = 2x\) and \(v' = -\sin x\). Apply the product rule to get the derivative: \(\frac{d}{dx}(x^2 \cos x) = 2x \cos x - x^2 \sin x\).

Step 2: Differentiate the Second Term

Now, differentiate the second term, \(4 \sin x\). The derivative of \(\sin x\) is \(\cos x\), so \(\frac{d}{dx}(4 \sin x) = 4 \cos x\).

Step 3: Find First Derivative

Combine the derivatives from Steps 1 and 2 to find the first derivative of \(y\): \(\frac{dy}{dx} = 2x \cos x - x^2 \sin x + 4 \cos x\).

Step 4: Differentiate Again for Second Derivative

Now, differentiate \(\frac{dy}{dx}\) to find \(\frac{d^2y}{dx^2}\). Differentiate each term separately. 1. For \(2x \cos x\), use the product rule: \(u' = 2\) and \(v' = -\sin x\), hence \(\frac{d}{dx}(2x \cos x) = -2x \sin x - 2 \cos x\).2. For \(-x^2 \sin x\), use the product rule: \(u' = 2x\) and \(v' = \cos x\), hence \(\frac{d}{dx}(-x^2 \sin x) = -2x \sin x - x^2 \cos x\).3. For \(4 \cos x\), the derivative is \(-4 \sin x\). Combine these: \(\frac{d^2y}{dx^2} = (-2x \sin x - 2 \cos x) + (-2x \sin x - x^2 \cos x) - 4 \sin x\).

Step 5: Simplify the Second Derivative

Combine like terms from Step 4 to simplify:- For terms involving \(-2x \sin x\), combine & simplify: \(-2x \sin x - 2x \sin x = -4x \sin x\).- The \(-2 \cos x\) term stands alone.- Combine \(-4 \sin x\) with above terms.- The term \(-x^2 \cos x\) also stands alone. Thus, \(\frac{d^2y}{dx^2} = -4x \sin x - 2 \cos x - x^2 \cos x - 4 \sin x\).

Key Concepts

Product Rule

The product rule is a crucial tool in calculus when dealing with derivatives of products of functions. When you have a function composed of two distinct terms, say \(u(x)\) and \(v(x)\), and you need to find the derivative of their product, the product rule becomes essential. Its formula is expressed as:\[\frac{d}{dx}(uv) = u'v + uv'\]

• \(u\) is the first function in the product. For example, in \(x^2 \cos x\), \(u = x^2\).
• \(v\) is the second function. In the same example, \(v = \cos x\).
• \(u'\) and \(v'\) are the derivatives of \(u\) and \(v\). These are calculated independently first.

Adding these derivatives using the product rule is a step-by-step process. It allows us to break down the complex interactions between functions into manageable pieces.
It's highly effective for functions like \(x^2 \cos x\), where the roles of each part are clear, and we can systematically calculate the derivative of the entire expression.

Differentiation

Differentiation is the fundamental process in calculus to find the rate at which a quantity changes. When differentiating a function, you determine its derivative, which gives insights about the curve's slope at any point. In the exercise, differentiation is applied twice, which is typical for finding the second derivative, such as \(\frac{d^2y}{dx^2}\).In this problem's context, we start by finding the first derivative of the function \(y = x^2 \cos x + 4 \sin x\). Then, by differentiating this first derivative, we obtain the second derivative:

• First Derivative: This is done by applying differentiation rules, like the product rule and standard derivatives of basic functions.
• Second Derivative: It involves a repeat application of differentiation on the first derivative to further specify how the rate of change itself is changing.

For our specific exercise, once we have the first derivative as a combination of differentiated terms, we continue by revisiting these terms, applying the respective rules again, and combining their derivatives. This step-by-step approach leads us to a precise expression describing the second rate of change in \(y\).

Trigonometric Functions

Trigonometric functions, such as \(\sin x\) and \(\cos x\), are a central part of many calculus problems, particularly those involving periodic behavior. These functions have specific differentiation rules that simplify the process of finding derivatives:

• Derivative of \(\sin x\): When differentiating \(\sin x\), it becomes \(\cos x\).
• Derivative of \(\cos x\): When differentiating \(\cos x\), it becomes \(-\sin x\).

These rules are employed seamlessly in solving our exercise. For example, in solving for \(\frac{d}{dx}(4 \sin x)\), we only switch \(\sin x\) to \(\cos x\), resulting in \(4 \cos x\).
Trigonometric derivatives also interact with algebraic elements from the product rule in our exercise. They provide a cyclical element since the derivatives repeat every few steps, reflecting the periodic nature of sine and cosine functions.