How do you differentiate f(x)=(3/(5x)-1)/(2/(x^2+7) using the quotient rule?

Dec 6, 2016

$f ' \left(x\right) = - \frac{10 {x}^{3} - 3 {x}^{2} + 21}{10 {x}^{2}}$

Explanation:

Let's start by converting this to the from $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$.

$f \left(x\right) = \frac{\frac{3 - 5 x}{5 x}}{\frac{2}{{x}^{2} + 7}}$

$f \left(x\right) = \frac{\left(3 - 5 x\right) \left({x}^{2} + 7\right)}{10 x}$

$f \left(x\right) = \frac{3 {x}^{2} - 5 {x}^{3} - 35 x + 21}{10 x}$

The derivative of a function $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$ is given by color(red)(f'(x) = (g'(x) * h(x) - h'(x)g(x))/(h(x))^2

$f ' \left(x\right) = \frac{\left(- 15 {x}^{2} + 6 x - 35\right) \left(10 x\right) - 10 \left(3 {x}^{2} - 5 {x}^{3} - 35 x + 21\right)}{10 x} ^ 2$

$f ' \left(x\right) = \frac{- 150 {x}^{3} + 60 {x}^{2} - 350 x - 30 {x}^{2} + 50 {x}^{3} + 350 x - 210}{100 {x}^{2}}$

$f ' \left(x\right) = \frac{- 100 {x}^{3} + 30 {x}^{2} - 210}{100 {x}^{2}}$

$f ' \left(x\right) = \frac{- 10 \left(10 {x}^{3} - 3 {x}^{2} + 21\right)}{100 {x}^{2}}$

$f ' \left(x\right) = - \frac{10 {x}^{3} - 3 {x}^{2} + 21}{10 {x}^{2}}$

Hopefully this helps!