Question

Find the derivatives of each of the functions in Exercises 17–50. In some cases it may be convenient to do some preliminary algebra

$f\left(x\right)=\frac{\sin \left(\arcsin x\right)}{\arctan x}$

Short answer

The derivative is$\frac{1}{\arctan x}-\frac{x}{\left(1+x^2\right){\arctan }^{2}x}$.

Step-by-step solution

Step 1. Given Information.

The given function is$f\left(x\right)=\frac{\sin \left(\arcsin x\right)}{\arctan x}$

Step 2. Preliminary Algebra.

We know,

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

Step 3. Derivative of the function.

The derivative of the function is,

$$\begin{gathered} \frac{d}{dx}\left(\frac{\sin \left(\arcsin x\right)}{\arctan x}\right)=\frac{d}{dx}\left(\frac{x}{\arctan x}\right) \\ =\frac{\left(\frac{d}{dx}x\right)\arctan x-x\left(\frac{d}{dx}\arctan x\right)}{(\arctan x{)}^2} \\ =\frac{\arctan x-\left(x\right)\frac{1}{1+x^2}}{(\arctan x{)}^2} \\ =\frac{\arctan x-\frac{x}{1+x^2}}{(\arctan x{)}^2} \\ =\frac{1}{\arctan x}-\frac{x}{\left(1+x^2\right){\arctan }^{2}x} \end{gathered}$$