Question

Find the derivatives of each of the functions in Exercises 51–62. In some cases it may be convenient to do some preliminary algebra. (These exercises involve hyperbolic functions and their inverses.)

$f\left(x\right)=\sin \left(e^{{\sinh }^{-1}x}\right)$

Short answer

The derivative is$\cos \left(e^{{\sinh }^{-1}x}\right)\left(\frac{e^{{\sinh }^{-1}x}}{\sqrt{x^2+1}}\right)$.

Step-by-step solution

Step 1. Given information.

The given function is$f\left(x\right)=\sin \left(e^{{\sinh }^{-1}x}\right)$.

Step 2. Preliminary Algebra.

We know,

$$\begin{gathered} \frac{d}{dx}\left(\sin x\right)=\cos x \\ \frac{d}{dx}\left({\sinh }^{-1}x\right)=\frac{1}{\sqrt{x^2+1}} \\ \frac{d}{dx}{e}^x=e^x \\ (f\circ u{)}^{\text{'}}(x)=f^{\text{'}}(u\left(x\right)\left)u^{\text{'}}\right(x) \end{gathered}$$

Step 3. Derivative.

The derivative of the function is,

$$\begin{gathered} \frac{d}{dx}\sin \left(e^{{\sinh }^{-1}x}\right)=\cos \left(e^{{\sinh }^{-1}x}\right)\left(\frac{d}{dx}{e}^{{\sinh }^{-1}x}\right) \\ =\cos \left(e^{{\sinh }^{-1}x}\right)\left(e^{{\sinh }^{-1}x}\frac{d}{dx}{\sinh }^{-1}x\right) \\ =\cos \left(e^{{\sinh }^{-1}x}\right)\left(e^{{\sinh }^{-1}x}\frac{1}{\sqrt{x^2+1}}\right) \\ =\cos \left(e^{{\sinh }^{-1}x}\right)\left(\frac{e^{{\sinh }^{-1}x}}{\sqrt{x^2+1}}\right) \end{gathered}$$