Question

Find a function $f$that has the given derivative $f\text{'}$ and value f(c).

$f\text{'}\left(x\right)=\frac{1}{1-4x^2},f\left(0\right)=0$

Short answer

The anti derivative is$f\left(x\right)=\frac{1}{2}\tan h^{-1}2x$

Step-by-step solution

Step 1:Given information

The given expression is$f\text{'}\left(x\right)=\frac{1}{1-4x^2},f\left(0\right)=0$

Step 2:Simplification

Given

$f\text{'}\left(x\right)=\frac{1}{1-4x^2}$

We will use an intelligent guess-and-check method to find f

Since differentiating a power function one and derivative of $tan{h}^{-1}x$is

$\frac{1}{1-x^2}$

One can start with $f\left(x\right)=\tan h^{-1}\left(2x\right)$

$f\text{'}\left(x\right)=\frac{d}{dx}\left(\tan h^{-1}2x\right)$

$=\frac{1}{1-4x^2}\frac{d}{dx}\left(2x\right)$

$=\frac{2}{1-4x^2}$

Hence, we can guess the function as

$f\left(x\right)=\frac{1}{2}\tan h^{-1}2x+c$

Now, given $f\left(0\right)=0$

$\Rightarrow 0=\tan h^{-1}\left(0\right)+c$

$\Rightarrow c=0$

Therefore

$f\left(x\right)=\frac{1}{2}\tan h^{-1}2x$