Question

86-89. Second derivatives 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 \(y = x \cos x^{2}\) is \(\frac{d^2y}{dx^2} = -4x\sin x^2\left(1 - x^2\right)\).

Step-by-step solution

Find the first derivative

To find the first derivative \(\frac{dy}{dx}\), we use the product rule for the function \(y = x \cos x^{2}\). The product rule states that if \(y = u(x)v(x)\), then \(\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)\). In our case, \(u(x) = x\) and \(v(x) = \cos x^{2}\), so: $$\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) = \frac{d}{dx}\left(x\right)\left(\cos x^{2}\right) + x\frac{d}{dx}\left(\cos x^2\right)$$

Calculate the derivatives

Now, we need to calculate the derivatives: \(\frac{d}{dx}\left(x\right)\) and \(\frac{d}{dx}\left(\cos x^2\right)\). The derivative of \(x\) is \(1\). For the derivative of \(\cos x^2\), we use the chain rule, which states if \(y = f(g(x))\), then \(\frac{dy}{dx} = f'(g(x))g'(x)\). Here, \(f(x) = \cos x\) and \(g(x) = x^2\): $$\frac{d}{dx}\left(\cos x^2\right) = -\sin x^2\left(2x\right) = -2x\sin x^2$$ Now, we plug these derivatives back into the expression for \(\frac{dy}{dx}\): $$\frac{dy}{dx} = (1)(\cos x^2) + x(-2x\sin x^2) = \cos x^2 - 2x^2\sin x^2$$

Find the second derivative

To find the second derivative \(\frac{d^2y}{dx^2}\), we take the derivative of \(\frac{dy}{dx}\) with respect to \(x\). So, we need to calculate the derivatives: \(\frac{d}{dx}\left(\cos x^2\right)\) and \(\frac{d}{dx}\left(-2x^2\sin x^2\right)\). We have already calculated the derivative for \(\cos x^2\) in Step 2: $$\frac{d}{dx}\left(\cos x^2\right) = -2x\sin x^2$$ For the derivative of \(-2x^2\sin x^2\), we again use the product rule. Let \(u(x) = -2x^2\) and \(v(x) = \sin x^2\). Applying the product rule: $$\frac{d}{dx}\left(-2x^2\sin x^2\right) = \left(\frac{d}{dx}(-2x^2)\right)\left(\sin x^2\right) + \left(-2x^2\right)\left(-2x\sin x^2\right)$$ Now, calculate the derivatives: \(\frac{d}{dx}\left(-2x^2\right)\) and \(\frac{d}{dx}\left(\sin x^2\right)\). The derivative of \(-2x^2\) is \(-4x\). For the derivative of \(\sin x^2\), we use the chain rule again, with \(f(x) = \sin x\) and \(g(x) = x^2\): $$\frac{d}{dx}\left(\sin x^2\right) = \cos x^2\left(2x\right) = 2x\cos x^2$$ Plugging the derivatives back into the expression for the second derivative: $$\frac{d^2y}{dx^2} = \left(-4x\right)\left(\sin x^2\right) -2x^2\left(-2x\sin x^2\right) = -4x\sin x^2 + 4x^3\sin x^2$$

Simplify the second derivative

Finally, we simplify the expression for the second derivative: $$\frac{d^2y}{dx^2} = -4x\sin x^2\left(1 - x^2\right)$$ So, the second derivative of the given function \(y = x \cos x^{2}\) is: $$\frac{d^2y}{dx^2} = -4x\sin x^2\left(1 - x^2\right)$$

Key concepts

Product Rule

When you are dealing with the product of two functions, you employ the Product Rule to differentiate. The rule helps you find derivatives where you have a multiplication situation. Imagine your function is made up of two parts, like a duo dancing together, each contributing its own steps to the entire performance.

• For the function \( y = u(x)v(x) \), the product rule is used to find the derivative as \( \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) \).
• Think of it as adding two terms: First, the derivative of the first function multiplied by the second function, and second, the first function multiplied by the derivative of the second function.

In our case \( y = x \cos x^{2} \), the two functions are \( u(x) = x \) and \( v(x) = \cos x^{2} \). This means you have to differentiate both \( x \) and \( \cos x^{2} \), applying the rule to each part to get the total derivative. That’s where the Chain Rule comes into play for the latter part.

Chain Rule

The Chain Rule is another powerful differentiation tool for unlocking derivatives of composite functions. If you have a function inside another function, like a box within a box, the Chain Rule is how you unpack that to find the derivative.

• For a composite function \( y = f(g(x)) \), the Chain Rule states that the derivative is \( \frac{dy}{dx} = f'(g(x))g'(x) \).
• You apply the derivative of the outer function multiplied by the derivative of the inner function.

In our example of the derivative \( \frac{d}{dx}(\cos x^2) \), \( f(x) = \cos x \) and \( g(x) = x^{2} \). Thus, using the Chain Rule, you calculate the derivative as \( -\sin x^{2} \times 2x \). The Chain Rule breaks down complicated expressions into manageable, step-by-step parts.

Differentiation

Differentiation is the fundamental process in calculus for finding a function that represents the rate of change of one quantity with respect to another. It helps us decipher how one variable changes with a small, yet significant tweak in another variable.

• Instructors teach differentiation through a plethora of rules: Product Rule, Chain Rule, Quotient Rule, just to name a few.
• Each rule serves a distinct purpose, adapting to different types of functions and combinations thereof.

In this exercise, you start with a function \( y = x \cos x^{2} \) and differentiate it step-by-step. First deriving \( \frac{dy}{dx} \) using the Product and Chain Rules, and then calculating \( \frac{d^{2}y}{dx^{2}} \) by differentiating again. Practical differentiation involves applying the correct rules thoroughly and then simplifying where possible to reach an understandable form.