Question

Prove each of the differentiation formulas in Exercises 93–96. (These exercises involve hyperbolic functions.)

$\frac{d}{dx}\left(\tan hx\right)=sec{h}^{2}x$

Short answer

We proved the formula$\frac{d}{dx}\left(\tan hx\right)=sec{h}^{2}x$.

Step-by-step solution

Step 1. Given Information

We need to prove the hyperbolic function $\frac{d}{dx}\left(\tan hx\right)=sec{h}^{2}x$.

Step 2. Proving the function 

$$\begin{gathered} \frac{d}{dx}\left(\tan hx\right) \\ =\frac{d}{dx}\left(\frac{\sin hx}{\cos hx}\right) \\ =\frac{d}{dx}\left(\frac{{\displaystyle \frac{e^x-e^{-x}}{2}}}{{\displaystyle \frac{e^x+e^{-x}}{2}}}\right) \\ =\frac{d}{dx}\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right) \\ =\frac{{\displaystyle \frac{d}{dx}\left(e^x-e^{-x}\right)\left(e^x+e^{-x}\right)-\left(e^x-e^{-x}\right)\frac{d}{dx}\left(e^x+e^{-x}\right)}}{{\left(e^x+e^{-x}\right)}^2} \\ =\frac{\left(e^x-\left(-e^{-x}\right)\right)\left(e^x+e^{-x}\right)-\left(e^x-e^{-x}\right)\left(e^x-e^{-x}\right)}{{\left(e^x+e^{-x}\right)}^2} \\ =\frac{\left(e^x+e^{-x}\right)\left(e^x+e^{-x}\right)-\left(e^x-e^{-x}\right)\left(e^x-e^{-x}\right)}{{\left(e^x+e^{-x}\right)}^2} \\ =\frac{{\left(e^x+e^{-x}\right)}^2-{\left(e^x-e^{-x}\right)}^2}{{\left(e^x+e^{-x}\right)}^2} \\ =\frac{e^{2x}+2e^x{e}^{-x}+e^{-2x}-e^{2x}+2e^{2x}{e}^{-x}-e^{-2x}}{{\left(e^x+e^{-x}\right)}^2} \\ =\frac{4e^x{e}^{-x}}{{\left(e^x+e^{-x}\right)}^2} \\ ={\left(\frac{2}{\left(e^x+e^{-x}\right)}\right)}^2 \\ =\frac{1}{\cos h^{2}x} \\ =sec{h}^{2}x \end{gathered}$$