Problem 35 - University Calculus: Early Transcendentals

Question

Find \(d y\). $$y=\tan ^{-1}\left(e^{x^{2}}\right)$$

Step-by-Step Solution

Answer

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

Step 1: Identify the Outer Function

The given function is a composition of functions, where the outer function is the inverse tangent, \(\tan^{-1}(u)\), where \(u = e^{x^2}\). We will first find the derivative of the outer function with respect to \(u\).

Step 2: Differentiate the Outer Function

Differentiate the outer function \(\tan^{-1}(u)\) with respect to \(u\). The derivative is \(\frac{1}{1+u^2}\).

Step 3: Identify the Inner Function

The inner function is \(u = e^{x^2}\). We will now find the derivative of this function with respect to \(x\).

Step 4: Differentiate the Inner Function

Differentiate \(u = e^{x^2}\) with respect to \(x\). Using the chain rule, the derivative is \(de^{x^2}/dx = e^{x^2} \cdot (2x)\).

Step 5: Apply the Chain Rule

Use the chain rule to differentiate \(y\) with respect to \(x\). That is, \(\frac{dy}{dx} = \frac{d}{du} [\tan^{-1}(u)] \cdot \frac{du}{dx}\). Substitute \(\frac{d}{du} [\tan^{-1}(u)] = \frac{1}{1+u^2}\) and \(\frac{du}{dx} = 2xe^{x^2}\).

Step 6: Simplify the Result

Substitute \(u = e^{x^2}\) back into the expression. The derivative becomes \(\frac{dy}{dx} = \frac{2xe^{x^2}}{1+(e^{x^2})^2}\). Therefore, \(dy = \frac{2xe^{x^2}}{1+e^{2x^2}}dx\).

Key Concepts

Chain Rule

The Chain Rule is a powerful tool in differential calculus used when differentiating composite functions. If you have a function composed of two or more functions, like our exercise where we dealt with \(y = \tan^{-1}(e^{x^2})\), you need the chain rule to differentiate it.

When using the chain rule, you follow these steps:

• First, identify your inner and outer functions. The function you see directly is the outer function, while the function inside, or nested, is the inner function.
• Next, differentiate the outer function with the inner function unchanged (treat it as a variable).
• Then, differentiate the inner function as it is.
• Finally, multiply the derivatives together to get your answer.

In our example, the outer function is \(\tan^{-1}(u)\) and the inner function is \(u = e^{x^2}\). We first found the derivative of \(\tan^{-1}(u)\) with respect to \(u\), resulting in \(\frac{1}{1+u^2}\), and then multiplied it by the derivative of inner function \(e^{x^2}\), which is \(2xe^{x^2}\). This gives us the entire derivative of the composite function.

Composite Functions

Composite functions occur when one function is nested within another. This is common in problems involving functions like trigonometric, exponential, or polynomial expressions.

Understanding composite functions is crucial because it allows us to deconstruct a complex expression into simpler parts. You recognize a composite function when you see one function "inside" another, like in our exercise where \(e^{x^2}\) is inside \(\tan^{-1}(u)\).
In terms of differentiation:

• The ‘outside’ part, or outer function, dictates what the function looks like overall.
• The ‘inside’ part, or inner function, determines the specific transformation of the variable within.

By breaking down the composite functions using these layers, and effectively using the Chain Rule, we simplify the differentiation process. This decomposition is evident in our example, where understanding how \(\tan^{-1}(e^{x^2})\) breaks into layers helps us apply the chain rule more easily.

Inverse Trigonometric Functions

Inverse trigonometric functions are used to find angles or values from trigonometric ratios. In our context, \(\tan^{-1}(e^{x^2})\) represents the angle whose tangent is \(e^{x^2}\).

These functions, including \(\sin^{-1}, \cos^{-1}, \tan^{-1}\) etc., have specific derivatives:

• The derivative of \(\tan^{-1}(x)\) is \(\frac{1}{1+x^2}\).
• These derivatives are useful when dealing with problems that involve finding the rate of change of angles concerning another variable.

In solving our differential calculus problem, knowing the derivative of \(\tan^{-1}(x)\) allows us to differentiate the outer function \(\tan^{-1}(e^{x^2})\) with respect to \(u\). Substituting back our function \(u = e^{x^2}\), we manage to connect the inverse trigonometric function with the composite function and complete the differentiation process. Understanding these functions helps us successfully handle calculus problems involving angles and inverses.