Problem 5 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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 1: Recall the definition of hyperbolic functions

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

Step 2: Find the inverse hyperbolic sine function

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

Step 3: 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: \(2x = e^y - e^{-y}\)

Step 4: Get rid of the negative exponent

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

Step 5: Create a common denominator

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

Step 6: Rewrite as a quadratic equation in \(e^y\)

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

Step 7: 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: \(e^y = x \pm \sqrt{x^2 + 1}\)

Step 8: 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: \(e^y = x + \sqrt{x^2 + 1}\)

Step 9: 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: \(y = \ln(x + \sqrt{x^2 + 1})\)

Step 10: Write the final expression for \(\sinh^{-1} x\)

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