How do you differentiate f(x)= (9-x)(5/(x^2 -4)) using the product rule?

May 26, 2018

$f ' \left(x\right) = 5 \cdot \frac{{x}^{2} - 18 x + 4}{{\left(x - 2\right)}^{2} \cdot {\left(x + 2\right)}^{2}}$

Explanation:

After the product rule
$\left(u v\right) ' = u ' v + u v '$
we get

$f ' \left(x\right) = - \frac{5}{{x}^{2} - 4} + \left(9 - x\right) \cdot \left(- 5\right) \cdot {\left({x}^{2} - 4\right)}^{- 2} \cdot 2 \cdot x$

which simplifies to
$f ' \left(x\right) = 5 \cdot \frac{{x}^{2} - 18 x + 4}{{\left(x - 2\right)}^{2} \cdot {\left(x + 2\right)}^{2}}$

May 26, 2018

$f ' \left(x\right) = \frac{5 \left({x}^{2} - 18 x + 4\right)}{{x}^{2} - 4} ^ 2$

Explanation:

$f \left(x\right) = \left(9 - x\right) \left(\frac{5}{{x}^{2} - 4}\right)$

Apply product rule,

$f ' \left(x\right) = \left[\frac{d}{\mathrm{dx}} \left(9 - x\right)\right] \left(\frac{5}{{x}^{2} - 4}\right) + \left[\frac{d}{\mathrm{dx}} \left(\frac{5}{{x}^{2} - 4}\right)\right] \left(9 - x\right)$

Differentiate inner terms,

$f ' \left(x\right) = \left(- 1\right) \left(\frac{5}{{x}^{2} - 4}\right) + \left(- \frac{10 x}{{x}^{2} - 4} ^ 2\right) \left(9 - x\right)$

Simplify,

$f ' \left(x\right) = - \frac{5}{{x}^{2} - 4} - \frac{10 x \left(9 - x\right)}{{x}^{2} - 4} ^ 2$

Common denominator,

$f ' \left(x\right) = - \frac{5 \left({x}^{2} - 4\right)}{{x}^{2} - 4} ^ 2 - \frac{10 x \left(9 - x\right)}{{x}^{2} - 4} ^ 2$

Simplify,

$f ' \left(x\right) = - \frac{5 {x}^{2} - 20}{{x}^{2} - 4} ^ 2 - \frac{90 x - 10 {x}^{2}}{{x}^{2} - 4} ^ 2$

Combine,

$f ' \left(x\right) = \frac{- 5 {x}^{2} + 20 - 90 x + 10 {x}^{2}}{{x}^{2} - 4} ^ 2$

Simplify,

$f ' \left(x\right) = \frac{5 {x}^{2} - 90 x + 20}{{x}^{2} - 4} ^ 2$

Factorise,

$f ' \left(x\right) = \frac{5 \left({x}^{2} - 18 x + 4\right)}{{x}^{2} - 4} ^ 2$