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.)

$f\text{'}\left(x\right)=\frac{3x}{1-9x^2}$

Short answer

The function that has the derivative $f\text{'}\left(x\right)=\frac{3x}{1-9x^2}$is $f\left(x\right)=-\frac{1}{6}\ln \left|1-9x^2\right|$.

Step-by-step solution

Step 1. Given Information.

The derivative:

$f\text{'}\left(x\right)=\frac{3x}{1-9x^2}$

Step 2. Guess the formula for the given derivative.

We know that,

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

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

Step 3. Check guess-and-check process.

Consider the function,

$$\begin{gathered} f\left(x\right)=\tan h^{-1}3x \\ f\text{'}\left(x\right)=\frac{d}{dx}\left(\tan h^{-1}3x\right) \\ =\frac{1}{1-{\left(3x\right)}^2}.\left(3\right) \\ =\frac{3}{1-9x^2} \end{gathered}$$

which is not a given derivative.

Step 4. Check another function.

Consider the function,

$$\begin{gathered} f\left(x\right)=-\frac{1}{6}\ln \left|1-9x^2\right| \\ f\text{'}\left(x\right)=-\frac{1}{6}.\frac{1}{1-9x^2}(-18x) \\ =\frac{3x}{1-9x^2} \end{gathered}$$

which is the given derivative.

So the function is $f\left(x\right)=-\frac{1}{6}\ln \left|1-9x^2\right|$.