How do you use the product Rule to find the derivative of f(x)= (9-x)(5/(x^2 -4))?

Aug 16, 2015

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

Explanation:

To differentiate this function using the product rule, you have to rewrite it as a product of two functions

$f \left(x\right) = {\underbrace{5}}_{\textcolor{\mathmr{and} a n \ge}{\text{constant}}} \cdot {\underbrace{\left(9 - x\right)}}_{\textcolor{b l u e}{g \left(x\right)}} \cdot {\underbrace{{\left({x}^{2} - 4\right)}^{- 1}}}_{\textcolor{g r e e n}{h \left(x\right)}}$

The product rule tells you that you can differentiate products of two functions by using the formula

color(blue)(d/dx(f(x)) = [d/dx(g(x))] * h(x) + g(x) * d/dx(h(x))

In your case, you would have

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = 5 \cdot \left(\left[\frac{d}{\mathrm{dx}} \left(9 - x\right)\right] \cdot {\left({x}^{2} - 4\right)}^{- 1} + \left(9 - x\right) \cdot \frac{d}{\mathrm{dx}} {\left({x}^{2} - 4\right)}^{- 1}\right)$

Now, to differentiate ${\left({x}^{2} - 4\right)}^{- 1}$, you can use the chain rule for ${u}^{- 1}$, with $u = \left({x}^{2} - 4\right)$. THis will get you

$\frac{d}{\mathrm{dx}} \left({u}^{- 1}\right) = \frac{d}{\mathrm{du}} \left({u}^{- 1}\right) \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

$\frac{d}{\mathrm{dx}} \left({u}^{- 1}\right) = - 1 \cdot {u}^{- 2} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} - 4\right)$

$\frac{d}{\mathrm{dx}} {\left({x}^{2} - 4\right)}^{- 1} = - {\left({x}^{2} - 4\right)}^{- 2} \cdot 2 x$

Take this back to your target derivative

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

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

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

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

You can write this as

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