Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(d y / d x\). \(y=\frac{1}{\sqrt[3]{x^{2} \sin x}}\)

Step-by-Step Solution

Answer

The derivative of the function is \( \frac{dy}{dx} = -\frac{1}{3} \left(\frac{2x \sin x + x^2 \cos x}{(x^2 \sin x)^{4/3}}\right) \).

Step 1: Rewrite the Function

Start by rewriting the given function in a form that can be more easily differentiated. The function is given as \( y = \frac{1}{\sqrt[3]{x^2 \sin x}} \). Rewrite it as a power function: \( y = (x^2 \sin x)^{-1/3} \).

Step 2: Differentiate Using the Chain Rule

We need to differentiate \( y = (x^2 \sin x)^{-1/3} \) with respect to \( x \). Apply the chain rule: \( \frac{dy}{dx} = -\frac{1}{3}(x^2 \sin x)^{-4/3} \cdot \frac{d}{dx}(x^2 \sin x) \).

Step 3: Differentiate the Inner Function

Now focus on differentiating the inner function. Given \( x^2 \sin x \), use the product rule: \[ \frac{d}{dx}(x^2 \sin x) = \frac{d}{dx}(x^2) \cdot \sin x + x^2 \cdot \frac{d}{dx}(\sin x) \]. This results in \( 2x \sin x + x^2 \cos x \).

Step 4: Combine Everything

Substitute the derivative of the inner function back into the expression from Step 2.So, \( \frac{dy}{dx} = -\frac{1}{3}(x^2 \sin x)^{-4/3}(2x \sin x + x^2 \cos x) \).

Step 5: Simplify the Expression

Simplify the derivative: \( \frac{dy}{dx} = -\frac{1}{3} \left(\frac{2x \sin x + x^2 \cos x}{(x^2 \sin x)^{4/3}}\right) \). This expression can be left as is for the final answer or further simplified based on context requirements.

Key Concepts

Chain Rule

The Chain Rule is a powerful technique in calculus used for differentiating compositions of functions. This occurs when you have a function inside another function. To apply the chain rule, you break down the differentiation process into steps. First, differentiate the outer function, leaving the inner function unchanged, and then multiply the result by the derivative of the inner function.

Consider the function given in the problem: \( y = (x^2 \sin x)^{-1/3} \). Here, the outer function is \( u^{-1/3} \) and the inner function is \( u = x^2 \sin x \). By applying the chain rule, the differentiation is executed as follows:

• Differentiate the outer part: \( \frac{d}{du}(u^{-1/3}) = -\frac{1}{3}u^{-4/3} \).
• Differentiate the inner part: \( \frac{d}{dx}(x^2 \sin x) \), as addressed in later steps.
• Combine both results: \( \frac{dy}{dx} = -\frac{1}{3}(x^2 \sin x)^{-4/3} \times \text{(derivative of the inner function)} \).

This allows you to handle complex functions by breaking them into simpler, manageable parts.

Product Rule

The Product Rule is essential for differentiating expressions where two functions are multiplied together. When you have a function that is the product of two functions, like \( g(x) = x^2 \) and \( h(x) = \sin x \) in this case, you need to use the product rule to find the derivative.

The product rule states that the derivative of \( f(x) = g(x)h(x) \) is given by:

• \( f'(x) = g'(x)h(x) + g(x)h'(x) \).

Let's apply this to our expression \( x^2 \sin x \):

• Differentiate \( x^2 \) to get \( 2x \).
• Differentiate \( \sin x \) to get \( \cos x \).
• Use the product rule: \( 2x \sin x + x^2 \cos x \).

The value \( 2x \sin x + x^2 \cos x \) is crucial as it forms the derivative of the inner function in the chain rule process.

Power Function

Power functions are expressions of the form \( x^n \), where \( n \) is any real number. They are simpler to differentiate compared to other functions because they follow a straightforward rule for differentiation.

The derivative of a power function \( x^n \) is given by:

• \( \frac{d}{dx}(x^n) = nx^{n-1} \).

In our original problem, the function \( (x^2 \sin x)^{-1/3} \) was initially presented as a cube root in a denominator. It was thus reformulated as a power function \( (x^2 \sin x)^{-1/3} \) for easier differentiation. This reformulation allows the application of the chain rule, as the power rule directly applies to the outer function \( u^{-1/3} \) in the composition.

Overall, rewriting complex functions into power functions is immensely helpful in calculus, making the application of differentiation rules more straightforward.