How do you integrate f(x)=(2x)/(x^3-7) using the quotient rule?

Nov 5, 2016

$f ' \left(x\right) = - \frac{4 {x}^{3} + 14}{{x}^{3} - 7} ^ 2$

Explanation:

The quotient rule states that if $f \left(x\right) = g \frac{x}{h \left(x\right)}$, then

$f ' \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2$.

So, where $f \left(x\right) = \frac{2 x}{{x}^{3} - 7}$, we see that:

$f ' \left(x\right) = \frac{\frac{d}{\mathrm{dx}} \left(2 x\right) \cdot \left({x}^{3} - 7\right) - 2 x \cdot \frac{d}{\mathrm{dx}} \left({x}^{3} - 7\right)}{{x}^{3} - 7} ^ 2$

Through the product rules: $\frac{d}{\mathrm{dx}} \left(2 x\right) = 2$ and $\frac{d}{\mathrm{dx}} \left({x}^{3} - 7\right) = 3 {x}^{2}$:

$f ' \left(x\right) = \frac{2 \left({x}^{3} - 7\right) - 2 x \left(3 {x}^{2}\right)}{{x}^{3} - 7} ^ 2$

$f ' \left(x\right) = \frac{2 {x}^{3} - 14 - 6 {x}^{3}}{{x}^{3} - 7} ^ 2$

$f ' \left(x\right) = - \frac{4 {x}^{3} + 14}{{x}^{3} - 7} ^ 2$