Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=\sinh ^{-1}\left(x^{2}\right) $$

Step-by-Step Solution

Answer

\( D_x y = \frac{2x}{\sqrt{x^4 + 1}} \)

Step 1: Understand the Expression

We are given the function \( y = \sinh^{-1}(x^2) \). The goal is to find the derivative of \( y \) with respect to \( x \), denoted as \( D_x y \). This means we need to differentiate the given function using appropriate rules.

Step 2: Recall Inverse Hyperbolic Sine Derivative Rule

Recall that the derivative of \( \sinh^{-1}(u) \) with respect to \( u \) is \( \frac{1}{\sqrt{u^2 + 1}} \). We will use this rule in combination with the chain rule, since our function \( u = x^2 \) is more complex than a single variable.

Step 3: Apply the Chain Rule

According to the chain rule, if a function \( y = f(u(x)) \) where \( u(x) = x^2 \), then the derivative \( \frac{dy}{dx} \) is given by \( \frac{dy}{du} \cdot \frac{du}{dx} \). Here, \( \frac{dy}{du} = \frac{1}{\sqrt{u^2 + 1}} \) and \( \frac{du}{dx} = 2x \).

Step 4: Differentiate \( u = x^2 \)

Calculate the derivative of \( u = x^2 \) with respect to \( x \). This derivative is \( \frac{du}{dx} = 2x \).

Step 5: Differentiate \( y = \sinh^{-1}(u) \)

Using the inverse hyperbolic sine derivative rule, the derivative of \( y = \sinh^{-1}(u) \) with respect to \( u \) is \( \frac{1}{\sqrt{u^2 + 1}} \). Substitute \( u = x^2 \), thus the derivative becomes \( \frac{1}{\sqrt{(x^2)^2 + 1}} = \frac{1}{\sqrt{x^4 + 1}} \).

Step 6: Combine the Results

Combine the results from Step 4 and Step 5 using the chain rule: \[\frac{dy}{dx} = \frac{1}{\sqrt{x^4 + 1}} \cdot 2x \].

Step 7: Express the Final Solution

Simplify the expression to find \( \frac{dy}{dx} = \frac{2x}{\sqrt{x^4 + 1}} \). This represents \( D_x y \).

Key Concepts

Inverse Hyperbolic Functions

Inverse hyperbolic functions are the inverse operations to hyperbolic functions. They are crucial in calculus because they allow us to solve equations involving hyperbolic functions. The three main inverse hyperbolic functions are:

• Inverse hyperbolic sine, denoted as \( ext{sinh}^{-1}(x) \)
• Inverse hyperbolic cosine, denoted as \( ext{cosh}^{-1}(x) \)
• Inverse hyperbolic tangent, denoted as \( ext{tanh}^{-1}(x) \)

These functions have their own derivative rules, which are essential when dealing with differentiation. For instance, the derivative of \( ext{sinh}^{-1}(u) \) with respect to \( u \) is \( \frac{1}{\sqrt{u^2 + 1}} \). It is important to remember this base derivative, especially when using the chain rule for more complex functions.

Chain Rule

The chain rule is an essential tool in calculus for finding the derivative of composite functions. When a function is composed of two functions, like \( y = f(u(x)) \), the chain rule allows us to differentiate it through a straightforward process. Here's how it works:

• First, differentiate the outer function with respect to the inner function: \( \frac{dy}{du} \).
• Then, differentiate the inner function with respect to \( x \): \( \frac{du}{dx} \).
• Finally, multiply the two derivatives: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).

By applying the chain rule, you can effectively manage complex differentiation tasks. In our exercise, we used it to find the derivative of \( y = \text{sinh}^{-1}(x^2) \) by considering \( x^2 \) as the inner function.

Differentiation Techniques

Differentiation techniques are diverse methods for finding derivatives and are vital for solving calculus problems. The choice of technique often depends on the form of the function. Common techniques include:

• Power rule: Used for simple polynomials.
• Product rule: Applies when multiplying two functions.
• Chain rule: Used for composite functions, as mentioned earlier.
• Implicit differentiation: Useful when functions are not explicitly solved for one variable.

Having a good grasp of these techniques allows you to tackle a variety of problems efficiently. In our example, the chain rule was the key technique used, enabling us to differentiate \( y = \text{sinh}^{-1}(x^2) \) step by step.

Derivative of Inverse Functions

The derivative of inverse functions plays a significant role in calculus, especially with functions that have inverses. An important rule for inverse functions is that if \( y = f^{-1}(x) \), then the derivative is:\[\frac{dy}{dx} = \frac{1}{f'(f^{-1}(x))}\]This rule underscores the relationship between a function and its inverse's derivatives. For inverse hyperbolic functions, we use their specific derivative rules tailored to their form. In our exercise, applying the derivative rule for \( \text{sinh}^{-1}(x) \) was an integral part of solving the problem. Understanding these derivatives helps to ensure correct and precise computation, especially when handling inverse relationships in calculus.