How do you differentiate f(x)= 1/(e^(3x) -x) using the quotient rule?

Apr 2, 2018

The derivative is $- \frac{3 {e}^{3 x} - 1}{{e}^{3 x} - x} ^ 2$.

Explanation:

Here's the quotient rule:

$\frac{d}{\mathrm{dx}} \left(f \frac{x}{g} \left(x\right)\right) \quad = \quad \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$

Let's use this in our problem:

$\textcolor{w h i t e}{=} \frac{d}{\mathrm{dx}} \left(\frac{1}{{e}^{3 x} - x}\right)$

$= \frac{\frac{d}{\mathrm{dx}} \left(1\right) \cdot \left({e}^{3 x} - x\right) - 1 \cdot \frac{d}{\mathrm{dx}} \left({e}^{3 x} - x\right)}{{e}^{3 x} - x} ^ 2$

$= \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{0 \cdot \left({e}^{3 x} - x\right)}}} - 1 \cdot \frac{d}{\mathrm{dx}} \left({e}^{3 x} - x\right)}{{e}^{3 x} - x} ^ 2$

$= \frac{- 1 \cdot \frac{d}{\mathrm{dx}} \left({e}^{3 x} - x\right)}{{e}^{3 x} - x} ^ 2$

$= - \frac{\frac{d}{\mathrm{dx}} \left({e}^{3 x}\right) - \frac{d}{\mathrm{dx}} \left(x\right)}{{e}^{3 x} - x} ^ 2$

$= - \frac{3 {e}^{3 x} - \frac{d}{\mathrm{dx}} \left(x\right)}{{e}^{3 x} - x} ^ 2$

$= - \frac{3 {e}^{3 x} - 1}{{e}^{3 x} - x} ^ 2$

That's it. Hope this helped!