Question

Find \(\frac{d^{2} y}{d x^{2}}\) for the following functions. $$y=x \cos x^{2}$$

Short answer

Answer: The second derivative of the function is \(\frac{d^2y}{dx^2} = -4x^3\cos x^2 - 4x\sin x^2 - \sin x^2\).

Step-by-step solution

Find the first derivative of the function

To find the first derivative, we will apply the product rule as the function \(y=x \cos x^2\) is a product of two functions (\(x\) and \(\cos x^2\)). The product rule states that if \(u(x)\) and \(v(x)\) are functions of \(x\), then $$\frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}.$$ In our case, let \(u=x\) and \(v=\cos x^2\). Then, \(\frac{du}{dx} = 1\) and \(\frac{dv}{dx} = -2x\sin x^2\) (by applying the chain rule). Applying the product rule, we get: $$\frac{dy}{dx} = x(-2x \sin x^2) + \cos x^2.$$

Find the second derivative of the function

Now that we have found the first derivative, let's find the second derivative by differentiating \(\frac{dy}{dx}\) with respect to \(x\). We will again apply the product rule, considering the term \((-2x^2 \sin x^2)\) as the product of two functions (\(-2x^2\) and \(\sin x^2\)). Let \(u=-2x^2\) and \(v=\sin x^2\). Then, \(\frac{du}{dx} = -4x\) and \(\frac{dv}{dx} = 2x\cos x^2\). Applying the product rule, we get: $$\frac{d^2y}{dx^2} = (-2x^2)(2x\cos x^2) + (\sin x^2)(-4x) + (-\sin x^2)$$ The second derivative of the function is: $$\frac{d^2y}{dx^2} = -4x^3\cos x^2 - 4x\sin x^2 - \sin x^2.$$

Key concepts

Product Rule

Understanding the product rule is essential when dealing with functions that are the product of two or more other functions. In calculus, this rule helps you find the derivative of such products. This method is particularly handy when functions are intertwined via multiplication.

The product rule states that if you have two functions, let's call them \( u(x) \) and \( v(x) \), then the derivative of their product \( uv \) with respect to \( x \) is:

• \[ \frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

In simple terms, it means you first take the derivative of the first function and multiply it by the second, then take the derivative of the second function and multiply it by the first. Add these two products together.

When applying this rule to the function \( y = x \cos x^2 \), you treat \( x \) as \( u \) and \( \cos x^2 \) as \( v \). You'll find the derivative of each and use the rule to combine them, which is a crucial step in finding how the entire product changes.

Chain Rule

The chain rule is a method for finding the derivative of composite functions. A composite function is essentially a function inside another function. Think of it like a nesting, where one layer needs to interact with another. This rule helps us differentiate such intricate constructions.

The chain rule is expressed as follows. If you have a composite function \( f(g(x)) \):

• \[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]

This means you need to differentiate the outer function \( f \) with respect to its inner function \( g(x) \), then multiply by the derivative of the inner function \( g \).

In the given exercise, the chain rule is used to differentiate \( \cos x^2 \). First, identify \( \cos(x^2) \) as the outer function, and the inner part \( x^2 \). Differentiate \( \cos \) with respect to \( x^2 \) (getting \( -\sin x^2 \)) and then the derivative of \( x^2 \) with respect to \( x \), which is \( 2x \). Multiply these to apply the chain rule effectively. This step is essential before applying the product rule in the context of this function.

Differentiation

Differentiation is a fundamental concept in calculus, providing a way to find the rate at which a function changes at any point. It involves computing derivatives, which essentially show how a function behaves and how it's likely to progress. This is integral to understanding the behavior of functions within an equation.

The process of differentiation takes complex expressions and simplifies them, revealing how one variable affects another. In our exercise, starting with the function \( y = x \cos x^2 \), differentiation is used to find both the first and second derivatives.

• The **first derivative** provides insight into the initial rate of change of the function with respect to \( x \).
• The **second derivative** gives information about the curvature of the function, essentially telling you how the rate of change itself is changing.

With the accurate application of product and chain rules, differentiation enables you to systematically break down and analyze the behavior of such functions.