Question

Theorem 7.8 a. The definition of the inverse hyperbolic cosine is \(y=\cosh ^{-1} x \Leftrightarrow x=\cosh y,\) for \(x \geq 1,0 \leq y<\infty .\) Use implicit differentiation to show that \(\frac{d}{d x}\left(\cosh ^{-1} x\right)=\frac{1}{\sqrt{x^{2}-1}}\) b. Differentiate \(\sinh ^{-1} x=\ln (x+\sqrt{x^{2}+1})\) to show that \(\frac{d}{d x}\left(\sinh ^{-1} x\right)=\frac{1}{\sqrt{x^{2}+1}}\)

Short answer

In this exercise, we found the derivatives of the inverse hyperbolic cosine and inverse hyperbolic sine functions using implicit differentiation. The results are: (a) The derivative of the inverse hyperbolic cosine function, \(\cosh^{-1} x\), is: \(\frac{dy}{dx} = \frac{1}{\sqrt{x^2 - 1}}\) (b) The derivative of the inverse hyperbolic sine function, \(\sinh^{-1} x\), is: \(\frac{d}{dx} = \frac{1}{\sqrt{x^2 + 1}}\)

Step-by-step solution

Write down the given equation

We are given the equation: \(y=\cosh^{-1}x \Leftrightarrow x=\cosh y\). We will use the second part of the bi-conditional expression: \(x=\cosh y\).

Apply the implicit differentiation

We will differentiate both sides with respect to x: \(\frac{d}{dx}x = \frac{d}{dx}\cosh y\) Using implicit differentiation, this gives: \(1 = \frac{dy}{dx} \sinh y\).

Solve for dy/dx

We need to solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{1}{\sinh y}\)

Express sinh y in terms of x

We know from our original equation that \(x = \cosh y\). Recalling the identity \(\cosh^2 y - \sinh^2 y = 1\), we can express \(\sinh y\) in terms of x: \(\sinh^2 y = \cosh^2 y - 1 = x^2 - 1\) Thus, \(\sinh y = \sqrt{x^2 - 1}\)

Substitute sinh y into dy/dx

Now we can substitute \(\sinh y\) back into our expression for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{1}{\sqrt{x^2 - 1}}\) This is the derivative of the inverse hyperbolic cosine function. Part (b):

Write down the given expression

We are given the equation: \(\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1})\).

Apply the differentiation

We will find the derivative of both sides with respect to x: \(\frac{d}{dx}\sinh^{-1} x = \frac{d}{dx} \ln(x + \sqrt{x^2 + 1})\) Using the chain rule, we get: \(\frac{d}{dx}\sinh^{-1} x = \frac{1}{x + \sqrt{x^2 + 1}} \cdot \frac{d}{dx}(x + \sqrt{x^2 + 1})\)

Differentiate the function inside the natural logarithm

Now we will differentiate the function inside the natural logarithm: \(\frac{d}{dx}(x + \sqrt{x^2 + 1}) = 1 + \frac{1}{2\sqrt{x^2 + 1}} \cdot 2x\) Simplifying this expression gives: \(\frac{d}{dx}(x + \sqrt{x^2 + 1}) = 1 + \frac{x}{\sqrt{x^2 + 1}}\)

Compute the final derivative

Now, substituting the expression back into our derivative: \(\frac{d}{dx}\sinh^{-1} x = \frac{1}{x + \sqrt{x^2 + 1}} \cdot \left(1 + \frac{x}{\sqrt{x^2+ 1}}\right)\) By simplifying the expression, we get: \(\frac{d}{dx}\sinh^{-1} x = \frac{1}{\sqrt{x^2+ 1}}\) This is the derivative of the inverse hyperbolic sine function.

Key concepts

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly solved for one variable in terms of another. This approach is particularly useful when dealing with inverse functions. In the case of our exercise, we're looking at \(x = \cosh y\). To find the derivative of an inverse hyperbolic function, we use implicit differentiation.
Assume \(y = \cosh^{-1} x\) and therefore \(x = \cosh y\). To differentiate both sides with respect to \(x\), apply the derivative operator to both sides: \(\frac{d}{dx} x = \frac{d}{dx} \cosh y\).
This results in \(1 = \sinh y \cdot \frac{dy}{dx}\), where we have applied the derivative of \(\cosh y\) and used the chain rule. Solving for \(\frac{dy}{dx}\), the result is \(\frac{1}{\sinh y}\). This leads to the expression for the derivative of interest.

Derivative of Inverse Hyperbolic Cosine

The derivative of the inverse hyperbolic cosine, denoted as \(\frac{d}{dx} (\cosh^{-1} x)\), is found using the implicit differentiation technique.
Starting from \(x = \cosh y\), and distinguishing it implicitly as shown, you reach \(\frac{dy}{dx} = \frac{1}{\sinh y}\).
The next step involves using the identity \(\cosh^2 y - \sinh^2 y = 1\) to express everything in terms of \(x\). Since \(x = \cosh y\), \(\sinh^2 y = x^2 - 1\), hence \(\sinh y = \sqrt{x^2 - 1}\).

• Substitute \(\sinh y\) back into the derivative: \(\frac{dy}{dx} = \frac{1}{\sqrt{x^2 - 1}}\).

This finds the derivative of the inverse hyperbolic cosine function, confirming its result.

Derivative of Inverse Hyperbolic Sine

To differentiate the inverse hyperbolic sine function, \(\sinh^{-1} x\), start by writing it as \(\ln(x + \sqrt{x^2 + 1})\).
Applying the chain rule, differentiate the logarithmic expression with respect to \(x\): \(\frac{d}{dx} \ln(x + \sqrt{x^2 + 1})\).
This results in: \(\frac{1}{x + \sqrt{x^2 + 1}} \cdot \frac{d}{dx}(x + \sqrt{x^2 + 1})\), where the chain rule is at play.

• Differentiating \(x + \sqrt{x^2 + 1}\) gives \(1 + \frac{x}{\sqrt{x^2 + 1}}\).
• Hence, plug this into the ongoing expression: \(\frac{d}{dx}\sinh^{-1} x = \frac{1}{x + \sqrt{x^2 + 1}}(1 + \frac{x}{\sqrt{x^2 + 1}})\).

Simplifying this, the resulting expression becomes \(\frac{1}{\sqrt{x^2+ 1}}\), establishing the derivative of the inverse hyperbolic sine function.

Chain Rule in Calculus

The chain rule is a fundamental technique in calculus used to differentiate composite functions. It allows us to differentiate a complex expression by breaking it down into simpler parts. When dealing with functions inside of other functions, the chain rule is a key component.
Consider the differentiation of \(\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1})\). Using the chain rule here means taking derivatives of nested functions, like \(\sqrt{x^2 + 1}\).

• Apply the chain rule: differentiate the outer function \(\ln(x + \sqrt{x^2 + 1})\) first.
• Then handle the inner part \(x + \sqrt{x^2 + 1}\), breaking it into \(1 + \frac{x}{\sqrt{x^2 + 1}}\).

This systematic approach is vital as it allows the calculation of derivatives that are not straightforward, such as those found in inverse hyperbolic function differentiation.