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

$f\left(x\right)={\tan }^{-1}\left(\ln x\right)$

Short answer

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

Step-by-step solution

Step 1. Given Information.

The given function is$f\left(x\right)={\tan }^{-1}\left(\ln x\right)$.

Step 2. Preliminary Algebra.

We know,

$$\begin{gathered} ChainRule, \\ (f\circ u{)}^{\text{'}}(x)=f^{\text{'}}(u\left(x\right)\left)u^{\text{'}}\right(x) \\ Also, \\ \left[\frac{d}{dx}\ln x=\frac{1}{x}\right] \end{gathered}$$

Step 3. Derivative of the function.

The derivative of the function,

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