Problem 21 - Calculus - AP Edition

Question

Find \(d y / d x\). $$y=e^{x \tan x}$$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = e^{x \tan x} (\tan x + x \sec^2 x) \).

Step 1: Identify the Differentiation Technique

To find \( \frac{dy}{dx} \) for the function \( y = e^{x \tan x} \), we observe the function is of the form \( e^{u(x)} \), where \( u(x) = x \tan x \). This form requires the use of the chain rule. We need to differentiate the exponential function and then multiply by the derivative of the exponent.

Step 2: Differentiate the Exponential Function

The derivative of \( e^{u(x)} \) with respect to \( x \) is \( e^{u(x)} \) multiplied by \( \frac{du}{dx} \). Therefore, the derivative at this step is: \[\frac{dy}{dx} = e^{x \tan x} \cdot \frac{d}{dx}(x \tan x).\]

Step 3: Differentiate the Exponent

Now, we need to differentiate the exponent \( x \tan x \). To do this, use the product rule, \((uv)' = u'v + uv'\), where \( u = x \) and \( v = \tan x \).* The derivative of \( u = x \) is \( u' = 1 \).* The derivative of \( v = \tan x \) is \( v' = \sec^2 x \).Applying the product rule: \[\frac{d}{dx}(x \tan x) = 1 \cdot \tan x + x \cdot \sec^2 x = \tan x + x \sec^2 x.\]

Step 4: Combine the Results

Substitute \( \frac{d}{dx}(x \tan x) = \tan x + x \sec^2 x \) back into the expression for \( \frac{dy}{dx} \): \[\frac{dy}{dx} = e^{x \tan x} \cdot (\tan x + x \sec^2 x).\] This represents the derivative of \( y \) with respect to \( x \).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate compositions of functions. When you have a function inside another function, like in \( y = e^{x \tan x} \), you can't simply apply the power rule or the derivative rule directly.
Instead, the chain rule allows you to understand how two different rates of change relate to each other.

• Inner function: Here, \( u(x) = x \tan x \) is the inner function.
• Outer function: The outer function is \( e^{u(x)} \).

To differentiate \( y \), we start by differentiating the outer function \( e^{u(x)} \) while keeping \( u(x) \) unchanged. The derivative is \( e^{u(x)} \).
However, because \( u(x) \) is also a function of \( x \), the chain rule requires us to multiply by \( \frac{du}{dx} \), the derivative of the inner function. This step ensures we account for how changes in \( x \) affect \( u(x) \).
In simple terms, the chain rule helps us "chain" these rates of change together correctly.

Product Rule

The product rule is another important differentiation technique used when differentiating products of two functions. The function \( u(x) = x \tan x \) takes the form of a product, requiring the product rule.
According to the product rule:

• If you have two functions \( u \) and \( v \), then the derivative of their product is \( (uv)' = u'v + uv' \).

In our case:

• Function \( u \): \( u = x \)
• Function \( v \): \( v = \tan x \)
• The derivatives are \( u' = 1 \) and \( v' = \sec^2 x \).

The product rule gives us the derivative of \( x \tan x \, \) which can be calculated as: \[ \frac{d}{dx}(x \tan x) = 1 \cdot \tan x + x \cdot \sec^2 x = \tan x + x \sec^2 x. \] Using the product rule helps simplify the process of finding the derivative of more complex functions that involve multiplicative relationships.

Derivatives of Trigonometric Functions

Differentiating trigonometric functions is a key skill in calculus, especially when combined with other functions. In this exercise, we dealt with \( \tan x \), a common trigonometric function.
The derivative of \( \tan x \) is \( \sec^2 x \).
This derivative arises from the identity \( \tan x = \frac{\sin x}{\cos x} \) and applying the quotient rule. However, once memorized, it's much more efficient to recall that \( \frac{d}{dx}(\tan x) = \sec^2 x \).
This result is particularly useful when using the product rule, as in differentiating \( x \tan x \).
Understanding derivatives of trigonometric functions is crucial for solving more advanced problems in calculus. They frequently appear in various forms, requiring a strong grasp of their rules to simplify and solve differential problems efficiently.