Problem 106 - Calculus Early Transcendentals

Question

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 \(\frac{d}{d x}\left(\cosh ^{-1} x\right)=\) \(1 / \sqrt{x^{2}-1}\). b. Differentiate \(\sinh ^{-1} x=\ln (x+\sqrt{x^{2}+1})\) to show that \(\frac{d}{d x}\left(\sinh ^{-1} x\right)=1 / \sqrt{x^{2}+1}\).

Step-by-Step Solution

Answer

Answer: The derivatives are as follows: - The derivative of the inverse hyperbolic cosine function: \(\frac{d}{dx}(\cosh^{-1} x) = \frac{1}{\sqrt{x^2 - 1}}\) - The derivative of the inverse hyperbolic sine function: \(\frac{d}{dx}(\sinh^{-1} x) = \frac{1}{\sqrt{x^2 + 1}}\)

Step 1: Differentiate the inverse hyperbolic cosine implicitly

Begin by differentiating both sides of the equation \(x = \cosh y\) with respect to x: \(\frac{d}{dx}(x) = \frac{d}{dx}(\cosh y)\) Since the left side is a simple derivative, we have: \(1 = \frac{d}{dx}(\cosh y)\) Next, apply the chain rule to differentiate \(\cosh y\) with respect to x: \(1 = \frac{dy}{dx}\cdot\frac{d}{dy}(\cosh y)\) Now, we know that the derivative of \(\cosh y\) with respect to y is \(\sinh y\). Therefore, we have: \(1 = \frac{dy}{dx}\cdot\sinh y\)

Step 2: Solve for dy/dx

To find \(\frac{dy}{dx}\), the derivative of the inverse hyperbolic cosine, solve for \(\frac{dy}{dx}\) in the equation \(1 = \frac{dy}{dx}\cdot\sinh y\), so: \(\frac{dy}{dx} = \frac{1}{\sinh y}\) Since we know that \(x = \cosh y\), we can use the identity \(\cosh^2 y - \sinh^2 y = 1\) to substitute for \(\sinh y\): \(\sinh^2 y = \cosh^2 y - 1\) \(\sinh y = \sqrt{\cosh^2 y - 1}\) And now replace \(\cosh y\) with x: \(\sinh y = \sqrt{x^2 - 1}\) Finally, substitute this back into the equation for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{1}{\sqrt{x^2 - 1}}\) So, the derivative of the inverse hyperbolic cosine is: \(\frac{d}{dx}(\cosh^{-1} x) = \frac{1}{\sqrt{x^2 - 1}}\)

Step 3: Differentiate the inverse hyperbolic sine implicitly

In part b, we have to differentiate \(\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1})\). So, we begin by differentiating both sides of the equation with respect to x: \(\frac{d}{dx}(\sinh^{-1} x) = \frac{d}{dx}\left[\ln(x + \sqrt{x^2 + 1})\right]\) Next, apply the chain rule to differentiate the natural logarithm: \(\frac{d}{dx}(\sinh^{-1} x) = \frac{1}{x + \sqrt{x^2 + 1}}\cdot\frac{d}{dx}\left[x + \sqrt{x^2 + 1}\right]\) Now, differentiate the term inside the square root: \(\frac{d}{dx}\left[x + \sqrt{x^2 + 1}\right] = 1 + \frac{1}{2}\cdot\frac{2x}{\sqrt{x^2 + 1}} = 1 + \frac{x}{\sqrt{x^2 + 1}}\)

Step 4: Combine and simplify

Substitute the differentiation result from Step 3 back into the equation from Step 2: \(\frac{d}{dx}(\sinh^{-1} x) = \frac{1}{x + \sqrt{x^2 + 1}}\cdot\left(1 + \frac{x}{\sqrt{x^2 + 1}}\right)\) Multiply the numerator terms to simplify the expression: \(\frac{d}{dx}(\sinh^{-1} x) = \frac{1 + x\left(\sqrt{x^2 + 1}\right)^{-1}}{\sqrt{x^2 + 1}}\) So, the derivative of the inverse hyperbolic sine is: \(\frac{d}{dx}(\sinh^{-1} x) = \frac{1}{\sqrt{x^2 + 1}}\) In conclusion, \(\frac{d}{dx}\left(\cosh^{-1} x\right) = \frac{1}{\sqrt{x^2 - 1}}\) and \(\frac{d}{dx}\left(\sinh^{-1} x\right) = \frac{1}{\sqrt{x^2 + 1}}\).

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used when it’s challenging to solve equations for one variable explicitly. In implicit differentiation, we differentiate each term with respect to one variable typically denoted as 'x', and assume the other variable (often 'y') is a function of 'x'.

• Begin by differentiating both sides of an equation.
• Use the chain rule during differentiation, as y is a function of x.
• Don't forget the derivative of expressions with y will include \( \frac{dy}{dx} \).

This method is powerful in calculus for functions defined in implicit forms where writing explicitly as y in terms of x is difficult or impossible.

Inverse Hyperbolic Functions

Inverse hyperbolic functions, including inverse hyperbolic cosine \( y = \cosh^{-1} x \), naturally appear in calculus when solving equations involving hyperbolic functions. These functions, although less common, are used in real-world applications like engineering and physics.

• For example, the inverse hyperbolic cosine is defined like \( y = \cosh^{-1} x \Rightarrow x = \cosh y \).
• Inverse hyperbolic functions can be expressed using logarithms and square roots making them critical in integration and differentiation.
• Suppose you need \( \cosh^{-1}(x) \) derivative, it would interestingly resemble an arc \(\frac{1}{\sqrt{x^2 - 1}}\).
• This computation often requires simplification using identity such as \( \cosh^2 y - \sinh^2 y = 1 \).

Derivatives

Derivatives measure how a function changes as its input changes. They are vital in calculus for understanding the behavior of functions. Each derivative tells us about the slope of the tangent line to the graph of a function at a given point.

• Finding the derivative of inverse functions often requires implicit differentiation to handle more complex structures such as inverse hyperbolic functions.
• The derivative of \( \cosh^{-1} x \) becomes \( \frac{1}{\sqrt{x^2 - 1}} \) revealing how it behaves relative to changes in x.
• Dealing with \( \sinh^{-1} x \), its derivative \( \frac{1}{\sqrt{x^2 + 1}} \) provides insight into the growth in varying x inputs.

Mastering derivatives involves this examination of function slopes, rates of change, and approximation of function behavior near any given point.

Chain Rule

The chain rule is an essential derivative rule for calculus when one function is nested inside another. It helps in differentiating composite functions by providing a way to handle the complexity.

• The chain rule expresses that the derivative of the composite of two functions is the product of the derivative of each function.
• In the context of the inverse hyperbolic functions, applying the chain rule simplifies finding derivatives when one component depends implicitly on another.
• For instance, in \( x = \cosh y \), the chain rule helps calculate the derivative: \( \frac{dy}{dx} \cdot \sinh y = 1 \).
• This approach is crucial when differentiating more complex layers of functions beyond basic algebraic expressions.

Understanding and applying the chain rule effectively is a major milestone in calculus, essential for tackling real-world mathematical models.