Question

We have to find the function of given derivative

$f^{\mathrm{\prime }}\left(x\right)=\frac{1}{1+9x^2}$

Short answer

The given function is the answer of given derivative

$f\left(x\right)=\frac{1}{3}{\tan }^{-1}3x$

Step-by-step solution

Step 1.Given information

We have been given the derivative function

$f^{\mathrm{\prime }}\left(x\right)=\frac{1}{1+9x^2}$

Step 2.Finding the approximate function of derivative 

One can think of ${\tan }^{-1}x$as the denominator is suggesting about this function

$\begin{array}{r}f\left(x\right)={\tan }^{-1}3x\\ f\left(x\right)={\tan }^{-1}3x\\ f^{\mathrm{\prime }}\left(x\right)=\frac{1}{1+(3x{)}^2}\frac{d}{dx}\left(3x\right)\\ =\frac{3}{1+9x^2}\end{array}$

Step 3.Finding more appropriate function 

$f\left(x\right)=\frac{1}{3}{\tan }^{-1}\left(3x^{}\right)$

Here we have used product rule

$\begin{array}{r}f\left(x\right)=\frac{1}{3}{\tan }^{-1}3x\\ f^{\mathrm{\prime }}\left(x\right)=\frac{1}{3}\left(\frac{1}{1+(3x{)}^2}\frac{d}{dx}\left(3x\right)\right)\\ =\frac{1}{3}\left(\frac{3}{1+9x^2}\right)\\ =\frac{1}{1+9x^2}\end{array}$