# How do you differentiate f(x)=(4x-x^2)^(1/2)/x^2 using the chain rule?

Dec 14, 2016

f'(x)=(x-6)/(x^2(4x-x^2)^(1/2)

#### Explanation:

We will have to use the chain rule but before that we have to use the $\textcolor{b l u e}{\text{quotient rule}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder " "If " f(x)=(g(x))/(h(x))" then }}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\textcolor{b l a c k}{f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2}} |}}} \leftarrow \text{ quotient rule}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder " " If " f(x)=g(h(x))" then}}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{f ' \left(x\right) = g ' \left(h \left(x\right)\right) . h ' \left(x\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}} \leftarrow \text{ chain rule}$

differentiate using the $\textcolor{b l u e}{\text{quotient rule}}$

$g \left(x\right) = {\left(4 x - {x}^{2}\right)}^{\frac{1}{2}} \text{ and using the chain rule}$

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

$= \frac{1}{2} {\left(4 x - {x}^{2}\right)}^{- \frac{1}{2}} \left(4 - 2 x\right) = \frac{1}{2} {\left(4 x - {x}^{2}\right)}^{- \frac{1}{2}} .2 \left(2 - x\right)$

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

$\text{ and } h \left(x\right) = {x}^{2} \Rightarrow h ' \left(x\right) = 2 x$
$\text{-------------------------------------------------------------}$

$\Rightarrow f ' \left(x\right) = \frac{{x}^{2} \left(2 - x\right) {\left(4 x - {x}^{2}\right)}^{- \frac{1}{2}} - {\left(4 x - {x}^{2}\right)}^{\frac{1}{2}} .2 x}{x} ^ 4$

simplifying the numerator.

$= \frac{x {\left(4 x - {x}^{2}\right)}^{- \frac{1}{2}} \left[x \left(2 - x\right) - \left(4 x - {x}^{2}\right) .2\right)}{x} ^ 4$

$= \frac{x {\left(4 x - {x}^{2}\right)}^{- \frac{1}{2}} \left[2 x - {x}^{2} - 8 x + 2 {x}^{2}\right]}{x} ^ 4$

$= \frac{x {\left(4 x - {x}^{2}\right)}^{- \frac{1}{2}} \left({x}^{2} - 6 x\right)}{x} ^ 4 = \frac{\cancel{{x}^{2}} {\left(4 x - {x}^{2}\right)}^{- \frac{1}{2}} \left(x - 6\right)}{\cancel{{x}^{4}} {x}^{2}}$

$= \frac{x - 6}{{x}^{2} {\left(4 x - {x}^{2}\right)}^{\frac{1}{2}}}$