Question

Find the derivatives of each of the functions in Exercises 17–44. In some cases it may be convenient to do some preliminary algebra.

$f\left(x\right)=\frac{1}{2-e^{5x}}$

Short answer

The required answer is $\frac{5e^{5x}}{{\left(2-e^{5x}\right)}^2}$.

Step-by-step solution

Step 1. Given information.

The given function is:

$f\left(x\right)=\frac{1}{2-e^{5x}}$

Step 2. Find the derivative of the given function.

$f\left(x\right)=\frac{1}{2-e^{5x}}$

As we can see that the function $f\left(x\right)$is a quotient, so we begin by applying the quotient rule,

$$\begin{gathered} f\text{'}\left(x\right)=\frac{\left(2-e^{5x}\right)·{\displaystyle \frac{d}{dx}}\left(1\right)-1·{\displaystyle \frac{d}{dx}}\left(2-e^{5x}\right)}{{\left(2-e^{5x}\right)}^2} \\ =\frac{\left(2-e^{5x}\right)·0-{\displaystyle \frac{d}{dx}}2+{\displaystyle \frac{d}{dx}}\left(e^{5x}\right)}{{\left(2-e^{5x}\right)}^2} \\ =\frac{0+5e^{5x}}{{\left(2-e^{5x}\right)}^2} \\ =\frac{5e^{5x}}{{\left(2-e^{5x}\right)}^2} \end{gathered}$$