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

Jul 28, 2017

The derivative is $= \frac{\left(x {e}^{x} - 1 - 2 {x}^{3} - {e}^{x}\right)}{{x}^{3} - {e}^{x} - 1} ^ 2$

Explanation:

The quotient rule is

$\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{{v}^{2}}$

We have

$f \left(x\right) = \frac{x}{{x}^{3} - {e}^{x} - 1}$

Here,

$u \left(x\right) = x$, $\implies$ $u ' \left(x\right) = 1$

$v \left(x\right) = {x}^{3} - {e}^{x} - 1$, $\implies$, $v ' \left(x\right) = 3 {x}^{2} - {e}^{x}$

Therefore,

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

$= \frac{\left({x}^{3} - {e}^{x} - 1 - 3 {x}^{3} + x {e}^{x}\right)}{{x}^{3} - {e}^{x} - 1} ^ 2$

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