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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 \(\frac{d}{d x}\left(\sinh ^{-1} x\right)=1 / \sqrt{x^{2}+1}.\)

Short Answer

Expert verified
Question: Find the derivatives of the following functions: a) \(\frac{d}{dx}(\cosh^{-1} x)\) b) \(\frac{d}{dx}(\sinh^{-1} x)\) given that \(\sinh^{-1} x = \ln(x+\sqrt{x^2 + 1})\) Solution: a) We have found that \(\frac{d}{dx}(\cosh^{-1} x)=\frac{1}{\sqrt{x^2 - 1}}\). b) Using the given expression for \(\sinh^{-1} x\), we have found that \(\frac{d}{dx}(\sinh^{-1} x)=\frac{1}{\sqrt{x^2 + 1}}\).

Step by step solution

01

Part a: Differentiating the inverse hyperbolic cosine implicitly

Start by considering the equation \(x=\cosh y\). Differentiate both sides of this equation with respect to \(x\): \begin{align*} \frac{d}{dx}(x) &= \frac{d}{dx}(\cosh y) \\ 1 &= (\sinh y)\frac{dy}{dx}. \end{align*}
02

Part a: Solving for \(\frac{dy}{dx}\)

Now, we solve for the derivative \(\frac{dy}{dx}\) by dividing both sides by \(\sinh y\). Due to the properties of hyperbolic functions, \(\sinh^2 y = \cosh^2 y - 1\), so we can write $$ \frac{dy}{dx} = \frac{1}{\sinh y} = \frac{1}{\sqrt{\cosh^2 y - 1}}. $$ Since \(x = \cosh y\), we can replace \(\cosh y\) with \(x\) in the above expression, and we obtain $$ \frac{dy}{dx} = \frac{1}{\sqrt{x^2 - 1}}. $$ Thus, we have shown that \(\frac{d}{dx}(\cosh^{-1} x)=\frac{1}{\sqrt{x^2 - 1}}\).
03

Part b: Differentiating the given expression of inverse hyperbolic sine

Given the expression \(\sinh^{-1} x = \ln(x+\sqrt{x^2 + 1})\), we will differentiate this expression with respect to \(x\): $$ \frac{d}{dx}(\sinh^{-1} x) = \frac{d}{dx}\left(\ln(x+\sqrt{x^2 + 1})\right). $$ Using chain rule, we have: $$ \frac{d}{dx}(\sinh^{-1} x) = \frac{1}{x+\sqrt{x^2+1}} \cdot \left(1+\frac{x}{\sqrt{x^2+1}}\right)^{-1}. $$
04

Part b: Simplifying the calculated derivative

Now, we will simplify obtained expression: $$ \frac{d}{dx}(\sinh^{-1} x) = \frac{1}{(x+\sqrt{x^2+1})\cdot\frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}} = \frac{1}{x+\sqrt{x^2+1}} \cdot \frac{\sqrt{x^2+1}}{x+\sqrt{x^2+1}}. $$ On canceling the common terms in the numerator and the denominator, we get: $$ \frac{d}{dx}(\sinh^{-1} x) = \frac{1}{\sqrt{x^2+1}}. $$ Thus, we have shown that \(\frac{d}{dx}(\sinh^{-1} x)=\frac{1}{\sqrt{x^2 + 1}}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Implicit Differentiation
Implicit differentiation is a powerful technique used to find the derivative of a function when it’s not easily expressed as a single open statement, like when dealing with inverse hyperbolic functions. This method becomes particularly handy when you have equations involving both variables entangled in such a manner that solving for one in terms of the other gets cumbersome. Rather than explicitly solving the function, you differentiate the whole equation directly with respect to a given variable (usually "x"), treating the other variable as an implicit function of the given variable.

In our exercise, we employ implicit differentiation to show how the derivative of the inverse hyperbolic cosine function comes about. Starting with the equation given, like \(x = \cosh y\), you differentiate both sides with respect to \(x\), effectively using the chain rule. You’ll often find this approach simplifies more complex differentiation problems, particularly when variables interdepend in expressions, revealing the necessary derivatives without isolating a variable first.

This method is immensely useful in calculus, extending the ability to find derivatives to broader classes of functions, such as hyperbolic functions, that naturally include implicit relationships.
Derivatives
Derivatives play a crucial role in calculus, often referred to as the study of change. They measure how a function's output value changes as its input changes, represented as the slope of the function at a given point. Calculating derivatives is fundamental not only in mathematical theories but also in numerous practical applications ranging from physics to economics.

In the context of our problem, derivatives help us find how quickly the inverse hyperbolic functions \(\cosh^{-1} x\) and \(\sinh^{-1} x\) change concerning the variable \(x\). Through differentiation, we obtain specific formulas that facilitate understanding how these functions behave. For instance, we derive \(\frac{d}{dx}(\cosh^{-1} x)\) as \(\frac{1}{\sqrt{x^2 - 1}}\) using both implicit differentiation and the inherent properties of hyperbolic functions.

Understanding derivatives of such functions not only involves mechanical computation but also enhances comprehension of their dynamic properties, providing depth and insight into how expressions change over different values and domains.
Hyperbolic Identities
Hyperbolic identities form an essential part of analyzing and computing with hyperbolic functions, which are counterparts to the trigonometric functions. These identities, such as \(\sinh^2 y = \cosh^2 y - 1\), parallel those of trigonometry but are adapted for the hyperbolic context. These can greatly simplify complex expressions and derivatives involving hyperbolic functions.

When differentiating inverse hyperbolic functions, like in our exercise, knowing these identities allows for direct substitution that simplifies the expression. For the hyperbolic cosine derivative task, the identity was used to solve for \(\frac{dy}{dx}\) effectively. It’s notable that these identities carve the path for finding elegant solutions without resorting to cumbersome algebraic manipulation.

Mastering hyperbolic identities is crucial for anyone dealing with hyperbolic functions, as they offer shortcuts and practical techniques for simplifying derivatives and solving differential equations. They interlock with the properties of the functions themselves, adding a layer of sophistication to understanding inverse hyperbolic derivatives.

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