# How do you derive a function composed of a division and a multiplication? Do you use the quotient rule or the product rule?

## Ex: f(x) = ((x^2-5x^-1)/((5x)(-.5x)))^3 (7x^6+4x^6-5x^2)

Apr 3, 2018

#### Explanation:

Consider the given function $f \left(x\right) = {\left(\frac{{x}^{2} - 5 {x}^{-} 1}{\left(5 x\right) \left(- .5 x\right)}\right)}^{3} \left(7 {x}^{6} + 4 {x}^{6} - 5 {x}^{2}\right)$

We can split it into two functions $g \left(x\right) = {\left(\frac{{x}^{2} - 5 {x}^{-} 1}{\left(5 x\right) \left(- .5 x\right)}\right)}^{3}$ and $h \left(x\right) = 7 {x}^{6} + 4 {x}^{6} - 5 {x}^{2} = 11 {x}^{6} - 5 {x}^{2}$.

Note that using product rule $\frac{\mathrm{df}}{\mathrm{dx}} = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$n and we know that $h ' \left(x\right) = \frac{\mathrm{dh}}{\mathrm{dx}} = 66 {x}^{5} - 10 x$, we need to find $g ' \left(x\right)$.

For $g \left(x\right)$ we can use chain rule, as $g \left(x\right) = {\left(u \left(x\right)\right)}^{3}$ and hence

$\frac{\mathrm{dg}}{\mathrm{dx}} = 3 {\left(u \left(x\right)\right)}^{2} = 3 {\left(\frac{{x}^{2} - 5 {x}^{-} 1}{\left(5 x\right) \left(- .5 x\right)}\right)}^{2} \times \frac{\mathrm{du}}{\mathrm{dx}}$

and as $u \left(x\right) = \frac{{x}^{2} - 5 {x}^{-} 1}{\left(5 x\right) \left(- .5 x\right)} = - \frac{{x}^{2} - \frac{1}{x}}{2.5 {x}^{2}}$

Now for finding $\frac{\mathrm{du}}{\mathrm{dx}}$, we can either have $u = - \frac{1}{2.5} + \frac{1}{2.5 {x}^{3}} = - 0.4 + 0.4 {x}^{- 3}$, which gives us $\frac{\mathrm{du}}{\mathrm{dx}} = - \frac{1.2}{x} ^ 4$ or use quotient formula, as follows

$\frac{\mathrm{du}}{\mathrm{dx}} = - \frac{2.5 {x}^{2} \left(2 x + \frac{1}{x} ^ 2\right) - 5 x \left({x}^{2} - \frac{1}{x}\right)}{6.25 {x}^{4}}$

= $- \frac{5 {x}^{3} + 2.5 - 5 {x}^{3} + 5}{6.25 {x}^{4}}$

= $- \frac{7.5}{6.25 {x}^{4}} = - \frac{1.2}{x} ^ 4$

Hence $\frac{\mathrm{dg}}{\mathrm{dx}} = 3 {\left(\frac{{x}^{2} - 5 {x}^{-} 1}{\left(5 x\right) \left(- .5 x\right)}\right)}^{2} \times - \frac{1.2}{x} ^ 4$

and $\frac{\mathrm{df}}{\mathrm{dx}} = 3 {\left(\frac{{x}^{2} - 5 {x}^{-} 1}{\left(5 x\right) \left(- .5 x\right)}\right)}^{2} \times - \frac{1.2}{x} ^ 4 \left(11 {x}^{6} - 5 {x}^{2}\right) + \left(66 {x}^{5} - 10 x\right) {\left(\frac{{x}^{2} - 5 {x}^{-} 1}{\left(5 x\right) \left(- .5 x\right)}\right)}^{3}$

and then simplified as indicated above.