Section 5.4 Deriving Inverse Hyperbolic Trigonmetric Functions
This section presents materials that explain or enable or use the following standards
- Calculate exponential derivatives
- Integrate polynomial, trig, and/or exponential functions
Using the same same method used to derive formulas for inverse trigonometric functions we can also derive them for hyperbolic trigonometric functions.
Subsection 5.4.1 Derive Inverse Hyperbolic Sine
\begin{align*}
x = \amp \sinh y. \\
= \amp \frac{e^y-e^{-y}}{2} \\
= \amp \frac{e^y-e^{-y}}{2} \cdot \frac{e^y}{e^y} \\
= \amp \frac{e^{2y}-1}{2e^y}. \\
2xe^y = \amp e^{2y}-1. \\
e^{2y}-2xe^y-1 = \amp 0. \\
u^2-2xu-1 = \amp 0.\\
u = \amp \frac{-(-2x)+\sqrt{(-2x)^2-4(1)(-1)}}{2(1)} \\
= \amp \frac{2x+\sqrt{4x^2+4}}{2} \\
= \amp \frac{2x+2\sqrt{x^2+1}}{2} \\
= \amp x+\sqrt{x^2+1}.\\
e^y = \amp x+\sqrt{x^2+1}.\\
\ln(e^y) = \amp \ln(x+\sqrt{x^2+1}).\\
y = \amp \ln(x+\sqrt{x^2+1}).
\end{align*}
Why was the \(\pm\) left out of the quadratic solution? Because \(e^y\) is always positive, so that would not provide a real solution for us.
\begin{align*}
y = \amp \ln(x+\sqrt{x^2+1}).\\
y^\prime = \amp \frac{1}{x+\sqrt{x^2+1}} \cdot \left( 1+\frac{1}{2}(x^2+1)^{-1/2}(2x) \right)\\
= \amp \frac{1+x(x^2+1)^{-1/2}}{x+\sqrt{x^2+1}}\\
= \amp \frac{1+x(x^2+1)^{-1/2}}{x+\sqrt{x^2+1}} \cdot \frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}\\
= \amp \frac{\sqrt{x^2+1}+x}{(x+\sqrt{x^2+1})\sqrt{x^2+1}}\\
= \amp \frac{1}{\sqrt{x^2+1}}.
\end{align*}
Notice in what ways this is similar to the derivative of inverse sine function and in which ways it is different.
