Question

In Exercises 72–77, find a function f that has the given derivative f. In each case you can find the answer with an educated guess-and-check process. (Some of these exercises involve hyperbolic functions.)

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

Short answer

The function that has the derivative $f\text{'}\left(x\right)=\frac{2x}{\sqrt{1+4x^2}}$is $f\left(x\right)=\frac{1}{2}\sqrt{1+4x^2}$.

Step-by-step solution

Step 1. Given Information.

The derivative:

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

Step 2. Guess the formula for the given derivative.

We know that,

$\frac{d}{dx}\left(\sin h^{-1}x\right)=\frac{1}{\sqrt{1+x^2}}$

So we can conclude that the derivative will be of the form of hyperbolic sinefunction.

Step 3. Check guess-and-check process.

$$\begin{gathered} f\left(x\right)=\sin h^{-1}\left(2x\right) \\ f\text{'}\left(x\right)=\frac{1}{\sqrt{1+x^2}}.\left(2\right) \\ =\frac{2}{\sqrt{1+x^2}} \end{gathered}$$

which is not a given derivative.

Step 4. Check another function.

Consider the function,

$$\begin{gathered} f\left(x\right)=\frac{1}{2}\sqrt{1+4x^2} \\ f\text{'}\left(x\right)=\frac{d}{dx}(\frac{1}{2}{(1+4x^2))}^{\frac{1}{2}}f\text{'}(x)=\frac{1}{2}.[\frac{1}{2}{(1+4x^2)}^{-\frac{1}{2}}.8x] \\ =\frac{1}{2}[\frac{4x}{\sqrt{1+4x^2}}] \\ =\frac{2x}{\sqrt{1+4x^2}} \end{gathered}$$

which is the given derivative.

So the function is $f\left(x\right)=\frac{1}{2}\sqrt{1+4x^2}$