Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=x^{3} \tan ^{-1}\left(e^{x}\right) $$

Step-by-Step Solution

Answer

\(D_x y = 3x^2 \tan^{-1}(e^x) + \frac{x^3 e^x}{1+e^{2x}}\).

Step 1: Identify the Differentiation Techniques

To find \(D_x y\), consider the function given: \(y = x^3 \tan^{-1}(e^x)\). Notice that this is a product of two functions: \(u = x^3\) and \(v = \tan^{-1}(e^x)\). We will use the product rule for differentiation, which states that \((uv)' = u'v + uv'\).

Step 2: Differentiate the First Function

Differentiate \(u = x^3\) with respect to \(x\):\[u' = \frac{d}{dx}(x^3) = 3x^2.\]

Step 3: Differentiate the Second Function

Differentiate \(v = \tan^{-1}(e^x)\) with respect to \(x\), using the chain rule. Let \(w = e^x\), then \(v = \tan^{-1}(w)\):1. Differentiate \(\tan^{-1}(w)\) with respect to \(w\), \(v' = \frac{1}{1+w^2}\).2. Differentiate \(w = e^x\) with respect to \(x\), \(w' = e^x\).3. By the chain rule, \(v' = \frac{1}{1+(e^x)^2} \cdot e^x = \frac{e^x}{1+e^{2x}}.\)

Step 4: Apply the Product Rule

Substitute the derivatives into the product rule: \[(uv)' = u'v + uv'.\]This gives:\[D_x y = 3x^2 \cdot \tan^{-1}(e^x) + x^3 \cdot \frac{e^x}{1+e^{2x}}.\]

Step 5: Simplify the Expression

Combine and simplify the terms, if needed:\[D_x y = 3x^2 \tan^{-1}(e^x) + \frac{x^3 e^x}{1+e^{2x}}.\]The expression is already as simplified as possible in terms of common algebraic simplifications.

Key Concepts

Product Rule

The product rule is a fundamental rule used in calculus for differentiating products of two or more functions. When you have a function that is the product of two functions, say \( u(x) \) and \( v(x) \), the derivative isn't simply the product of their derivatives. Instead, the product rule tells us how to find the derivative:

• \((uv)' = u'v + uv'\)

To apply the product rule effectively, follow these steps:

• Differentiation: Start by differentiating each function separately. For instance, if \( u(x) = x^3 \), then \( u'(x) = 3x^2 \).


• Application: Use the formula to find the combined derivative. Multiply the derivative of the first function by the second function, then add it to the first function times the derivative of the second.

These principles are applied while solving the exercise, involving the differentiation of \( y = x^3 \tan^{-1}(e^x) \). The derivative, \( D_x y \), is computed by applying the product rule. First, differentiate \( x^3 \) to get \( 3x^2 \), and then differentiate \( \tan^{-1}(e^x) \) using the chain rule, before combining them using the product rule.

Chain Rule

The chain rule is crucial for finding derivatives of composite functions. It provides a way to differentiate a function that is composed of two or more functions. For a function where the input is itself a function, the chain rule is essential. Consider a composite function \( f(g(x)) \), the chain rule states:

• The derivative of \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \).

In our exercise, the chain rule helps differentiate \( \tan^{-1}(e^x) \). Here's how:

• Set \( w = e^x \); the function becomes \( \tan^{-1}(w) \).


• Differentiate \( \tan^{-1}(w) \) with respect to \( w \). The derivative is \( \frac{1}{1+w^2} \).


• Find the derivative of \( w = e^x \) concerning \( x \), which is \( e^x \).


• Apply the chain rule: Multiply these derivatives together to get \( \frac{e^x}{1+e^{2x}} \).

Understanding and applying the chain rule effectively enables us to handle even complex functions where the input is another function, as seen in the derivative calculation of \( \tan^{-1}(e^x) \).

Inverse Trigonometric Functions

Inverse trigonometric functions take trigonometric ratios and return the corresponding angles. They are vital in calculus and have special differentiation rules. For example, if you have \( y = \tan^{-1}(x) \), the derivative is:

• \( \frac{dy}{dx} = \frac{1}{1+x^2} \)

When inverse trigonometric functions are part of a composite function, use the chain rule. In the exercise with \( \tan^{-1}(e^x) \), the derivative involves differentiating \( \tan^{-1} \) and applying the derivative of the inner function \( e^x \).

These functions, especially \( \tan^{-1} \), appear frequently in various fields, providing solutions to equations where angles are not directly evident. Understanding how to differentiate them enables you to explore deeper into calculus problems involving angles and ratios effectively.