Question

Find a function f from its derivative

$f^{\mathrm{\prime }}\left(x\right)=\frac{3x}{1+9x^2}$

Short answer

The answer of given derivative function is

$f\left(x\right)=\frac{1}{6}\ln \left(1+9x^2\right)$

Step-by-step solution

Step 1.Given information

We have been given the derivative of function

$f^{\mathrm{\prime }}\left(x\right)=\frac{3x}{1+9x^2}$

Step 2. Finding the function 

One can suspect from the derivative that function${\tan }^{-1}x$

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

Step 3. Getting more approximate function 

Now the approximate function is$f\left(x\right)=\frac{1}{6}\ln \left(1+9x^2\right)$

$\begin{array}{r}f\left(x\right)=\frac{1}{6}\ln \left(1+9x^2\right)\\ f^{\mathrm{\prime }}\left(x\right)=\frac{1}{6}\frac{d}{dx}\ln \left(1+9x^2\right)\\ =\frac{1}{6\left(1+9x^2\right)}\frac{d}{dx}\left(1+9x^2\right)\left[\frac{d}{dx}\ln x=\frac{1}{x}\right]\\ =\frac{1}{6\left(1+9x^2\right)}\left(\frac{d}{dx}1+\frac{d}{dx}9x^2\right)\\ =\frac{9\left(2x\right)}{6\left(1+9x^2\right)}\\ =\frac{3x}{\left(1+9x^2\right)}\end{array}$