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Chapter 6: Problem 5

Express \(\sinh ^{-1} x\) in terms of logarithms.

Short Answer

Expert verified
Question: Express the inverse hyperbolic sine function, \(\sinh^{-1} x\), in terms of logarithms. Answer: \(\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1})\)

Step by step solution

01

Recall the definition of hyperbolic functions

By definition, the hyperbolic sine function is given by: \(\sinh x = \frac{e^x - e^{-x}}{2}\)
02

Find the inverse hyperbolic sine function

Let \(y = \sinh ^{-1} x\). Then, \(x = \sinh y\), which means: \(x = \frac{e^y - e^{-y}}{2}\)
03

Rewrite the equation in terms of exponentials

We need to rewrite the equation in Step 2 in terms of exponentials to make it more manageable. Multiply both sides of the equation by 2: \(2x = e^y - e^{-y}\)
04

Get rid of the negative exponent

To get rid of the negative exponent, rewrite \(e^{-y}\) as \(\frac{1}{e^y}\): \(2x = e^y - \frac{1}{e^y}\)
05

Create a common denominator

To simplify the equation further, create a common denominator by multiplying both terms on the right by \(e^y\): \(2x e^y = e^{2y} - 1\)
06

Rewrite as a quadratic equation in \(e^y\)

Now, rewrite the equation as a quadratic equation in \(e^y\): \(e^{2y} - 2x e^y - 1 = 0\)
07

Solve the quadratic equation

Let \(u = e^y\). Then, the quadratic equation becomes: \(u^2 - 2x u - 1 = 0\) To solve this quadratic equation, we'll use the quadratic formula. The formula is: \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) For our equation, we have \(a=1\), \(b=-2x\), and \(c=-1\). Plugging in these values, we get: \(u = \frac{2x \pm \sqrt{(-2x)^2 - 4(1)(-1)}}{2(1)}\) Simplify the expression: \(u = x \pm \sqrt{x^2 + 1}\) Here, \(u = e^y\), so we have: \(e^y = x \pm \sqrt{x^2 + 1}\)
08

Choose the appropriate solution

We have two possible solutions for \(e^y\): \(x + \sqrt{x^2 + 1}\) or \(x - \sqrt{x^2 + 1}\). Since \(e^y\) is always positive and the second solution could be negative, we choose the first solution: \(e^y = x + \sqrt{x^2 + 1}\)
09

Take the natural logarithm of both sides

Now, take the natural logarithm of both sides to solve for \(y\): \(\ln(e^y) = \ln(x + \sqrt{x^2 + 1})\) As \(\ln(e^y) = y\), we finally have: \(y = \ln(x + \sqrt{x^2 + 1})\)
10

Write the final expression for \(\sinh^{-1} x\)

Since \(y = \sinh^{-1} x\), the final expression is: \(\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1})\)

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Most popular questions from this chapter

There are several ways to express the indefinite integral of \(\operatorname{sech} x\). a. Show that \(\left.\int \operatorname{sech} x d x=\tan ^{-1}(\sinh x)+C \text { (Theorem } 6.9\right)\) (Hint: Write sech \(x=\frac{1}{\cosh x}=\frac{\cosh x}{\cosh ^{2} x}=\frac{\cosh x}{1+\sinh ^{2} x}\) and then make a change of variables.) b. Show that \(\int \operatorname{sech} x d x=\sin ^{-1}(\tanh x)+C .\) (Hint: Show that \(\operatorname{sech} x=\frac{\operatorname{sech}^{2} x}{\sqrt{1-\tanh ^{2} x}}\) and then make a change of variables.) c. Verify that \(\int \operatorname{sech} x d x=2 \tan ^{-1} e^{x}+C\) by proving \(\frac{d}{d x}\left(2 \tan ^{-1} e^{x}\right)=\operatorname{sech} x.\)

Verify the following identities. $$\cosh \left(\sinh ^{-1} x\right)=\sqrt{x^{2}+1}, \text { for all } x$$

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

Let $$f(x)=\left\\{\begin{array}{cl}x & \text { if } 0 \leq x \leq 2 \\\2 x-2 & \text { if } 2

The U.S. government reports the rate of inflation (as measured by the Consumer Price Index) both monthly and annually. Suppose that for a particular month, the monthly rate of inflation is reported as \(0.8 \%\). Assuming that this rate remains constant, what is the corresponding annual rate of inflation? Is the annual rate 12 times the monthly rate? Explain.
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