Problem 429 - Calculus Volume 1

Question

For the following exercises, solve each problem. Prove the formula for the derivative of \(y=\sinh ^{-1}(x)\) by differentiating \(x=\sinh (y) .\) (Hint: Use hyperbolic trigonometric identities.)

Step-by-Step Solution

Answer

The derivative of \( y = \sinh^{-1}(x) \) is \( \frac{1}{\sqrt{1 + x^2}} \).

Step 1: Establish the Initial Equation

We begin by considering the equation given in the problem: \( x = \sinh(y) \). The goal is to find the derivative of \( y = \sinh^{-1}(x) \), which involves differentiating \( y \) with respect to \( x \).

Step 2: Differentiate with Respect to x

Differentiate both sides of the equation \( x = \sinh(y) \) implicitly with respect to \( x \). We have: \( \frac{dx}{dx} = \cosh(y) \cdot \frac{dy}{dx} \). Since \( \frac{dx}{dx} = 1 \), we can rewrite this as \( 1 = \cosh(y) \cdot \frac{dy}{dx} \).

Step 3: Express \( \cosh(y) \) in Terms of x

Recall the hyperbolic identity \( \cosh^2(y) - \sinh^2(y) = 1 \). Substitute \( x = \sinh(y) \) into this identity to get \( \cosh^2(y) = 1 + x^2 \). Thus, \( \cosh(y) = \sqrt{1 + x^2} \).

Step 4: Solve for \( \frac{dy}{dx} \)

Substitute \( \cosh(y) = \sqrt{1 + x^2} \) into the previous equation from Step 2: \( 1 = \sqrt{1 + x^2} \cdot \frac{dy}{dx} \). Solve for \( \frac{dy}{dx} \) to obtain: \( \frac{dy}{dx} = \frac{1}{\sqrt{1 + x^2}} \).

Step 5: Conclusion

We have derived that the derivative of \( y = \sinh^{-1}(x) \) with respect to \( x \) is \( \frac{1}{\sqrt{1 + x^2}} \). This proves the formula for the derivative of the inverse hyperbolic sine function.

Key Concepts

Hyperbolic Trigonometric Identities

Hyperbolic trigonometric identities are essential when working with hyperbolic functions. Just like their trigonometric counterparts, hyperbolic functions have identities that help simplify and solve equations. Two of the main hyperbolic functions are the hyperbolic sine \(\sinh(y)\) and the hyperbolic cosine \(\cosh(y)\). Understanding these identities can make it easier to manipulate and solve equations involving hyperbolic functions.
One of the most important identities used in problems involving hyperbolic functions is \(\cosh^2(y) - \sinh^2(y) = 1\). This identity is similar to the Pythagorean identity in trigonometry \(\cos^2(\theta) + \sin^2(\theta) = 1\). By rearranging and using substitutions, we can express one function in terms of another. This often involves transforming expressions like \(\cosh(y)\) using known values of \(\sinh(y)\). In our problem, it's crucial to express \(\cosh(y)\) in terms of \(x\) because \(x = \sinh(y)\). Substituting \(x\) into the identity gives us \(\cosh(y) = \sqrt{1 + x^2}\).
These identities not only facilitate differentiation of hyperbolic functions but also help us switch from one form of a function to another.

Implicit Differentiation

Implicit differentiation is a technique you use when you have an equation involving a function and its inverse. When functions don't have explicit forms, like \(y = \sinh^{-1}(x),\) you need implicit differentiation to find derivatives.
This method requires you to treat one variable as a function of another. When you differentiate implicitly, you're essentially acknowledging that there is a hidden dependence of one variable on another and you differentiate both sides of the equation with respect to one variable.
In the given problem, you start with \(x = \sinh(y)\) and differentiate both sides with respect to \(x\). This requires you to use the chain rule, particularly when one side involves \(y\), which itself is a function of \(x\). You get \(\frac{dx}{dx} = \cosh(y) \cdot \frac{dy}{dx}\). Remember, our aim is to find \(\frac{dy}{dx}\).
By treating \(y\) as an implicit function of \(x\), we can rearrange to solve for \(\frac{dy}{dx}\). This is where implicit differentiation shines, especially when dealing with more complex inverse relationships that aren't easily isolated.

Inverse Functions

Inverse functions reverse the effect of a original function. In simpler terms, if you input a result back into an inverse function, you get back to the original input. The inverse of a function \(f(x)\) is typically denoted as \(f^{-1}(x)\). For hyperbolic functions, if \(x = \sinh(y)\), then the inverse function is \(y = \sinh^{-1}(x)\).
Why are inverse functions important in calculus? Because they allow us to find original input values given the output, and are essential when determining derivatives. Differentiating an inverse function involves finding \(\frac{dy}{dx}\), where \(y\) is the inverse function of \(x\).
In the exercise, you prove the derivative of the inverse hyperbolic sine function. The process incorporates both implicit differentiation and hyperbolic identities, enabling us to express one hyperbolic function in terms of another. This results in the derivative of \(y = \sinh^{-1}(x)\) being \(\frac{1}{\sqrt{1 + x^2}}\), demonstrating how inverse functions are pivotal in calculus for understanding rate of change and behavior of functions.