Question

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

Short answer

The derivative is \(3x^2 \tan^{-1}(e^x) + \frac{x^3 e^x}{1 + e^{2x}}\).

Step-by-step solution

Identify the Derivative Rules Needed

To find the derivative of the given function, identify the appropriate rules. In this case, we will need to use the product rule because the function is a product of two functions: \(x^3\) and \(\tan^{-1}(e^x)\). Additionally, the chain rule will be needed to differentiate \(\tan^{-1}(e^x)\).

Apply the Product Rule

The product rule states that if you have two functions, \(u\) and \(v\), then their derivative is \(u'v + uv'\). Let \(u = x^3\) and \(v = \tan^{-1}(e^x)\). Now, differentiate \(u\) to get \(u' = 3x^2\).

Differentiate \(v\) Using the Chain Rule

The derivative of \(v = \tan^{-1}(e^x)\) is found using the chain rule. Recall that \(\frac{d}{dx}\tan^{-1}(x) = \frac{1}{1 + x^2}\). Using the chain rule, \(v' = \frac{1}{1 + (e^x)^2} \cdot \frac{d}{dx}[e^x]\). Since \(\frac{d}{dx}[e^x] = e^x\), we have \(v' = \frac{e^x}{1 + e^{2x}}\).

Substitute and Simplify

Substitute \(u\), \(u'\), \(v\), and \(v'\) back into the product rule expression: \(u'v + uv'\). This gives:\[3x^2 \cdot \tan^{-1}(e^x) + x^3 \cdot \frac{e^x}{1 + e^{2x}}\]Simplify the expression to obtain the final derivative.

Key concepts

Product Rule

The product rule in calculus is a fantastic tool used when differentiating products of two functions. It simplifies the process by breaking it into manageable steps. Imagine you have two functions, let's call them \( u \) and \( v \), and you need the derivative of their product, \( uv \). The product rule teaches us that the derivative is not just a simple product of their individual derivatives. Instead, it follows the formula:

• \( (uv)' = u'v + uv' \)

This might seem confusing at first, but it breaks down as follows:- Differentiate the first function \( u \) while keeping \( v \) constant, multiply them to get \( u'v \).- Next, keep \( u \) constant while differentiating \( v \), resulting in \( uv' \).- Finally, add these two expressions together.
For the example \( y = x^3 \tan^{-1}(e^x) \), we let \( u = x^3 \) and \( v = \tan^{-1}(e^x) \). Differentiating \( u \) gives \( u' = 3x^2 \). The product rule guides us through these steps to then combine \( u' \) and \( v \) with \( u \) and \( v' \). It's like solving a puzzle, where each piece is necessary to complete the whole picture.

Chain Rule

The chain rule is an essential technique in calculus for finding the derivative of composite functions. These are functions within functions, kind of like layers. The rule helps you "peel back" each of these layers to discover the derivative of the entire expression. Here’s a breakdown of the chain rule:

• If a function \( y \) can be expressed as \( f(g(x)) \), the derivative \( y' \) can be found using:
• \( y' = f'(g(x)) \, g'(x) \)

This essentially means:- Differentiate the outer function \( f \) as if it were "normal," but leave the inner function \( g(x) \) untouched for now.- Multiply this result by the derivative of the inner function \( g(x) \).- Put it all together!
In the exercise, \( v = \tan^{-1}(e^x) \), we used the chain rule to find its derivative. First, consider the outer function \( \tan^{-1}(x) \), with a known derivative of \( \frac{1}{1 + x^2} \). Then apply this by substituting \( e^x \) for \( x \), and multiply by the derivative of \( e^x \), resulting in the formula \( \frac{e^x}{1 + (e^x)^2} \). This step-by-step method allows us to handle even the most complex nested functions with ease.

Inverse Trigonometric Functions

Inverse trigonometric functions are the "mirrors" of their regular trigonometric counterparts. They reverse the process of these functions, helping to find angles from given trigonometric ratios. In calculus, we often need to work with derivatives of these inverse functions.

• For example, the derivative of \( \tan^{-1}(x) \) is \( \frac{1}{1+x^2} \).

This formula is crucial when differentiating expressions like \( \tan^{-1}(e^x) \), as seen in the exercise. To use the inverse trigonometric derivative effectively, combine it with other derivative rules, such as the chain rule.
Inverse functions provide a way to tackle problems where you're given a result and need an input, instead of the other way around. They are essential in calculus for integrating and differentiating functions involving angles and are incredibly helpful in simplifying complex expressions involving products like in this exercise. By understanding these processes, you unlock a powerful method for addressing a wide range of calculus problems.