# How do you differentiate  f(x)=x/sqrt(3-xe^x) using the chain rule.?

Jan 15, 2016

$f ' \left(x\right) = \frac{1}{2 \sqrt{{\left(3 - x {e}^{x}\right)}^{3}}} \left({x}^{2} {e}^{x} - x {e}^{x} + 6\right)$

#### Explanation:

$f \left(x\right) = \frac{x}{\sqrt{3 - x {e}^{x}}} = \sqrt{{x}^{2} / \left(3 - x {e}^{x}\right)}$

Using the Chain Rule

given

$y = f \left(g \left(x\right)\right)$

we could think

$u = g \left(x\right)$; $y = f \left(u\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} f \left(g \left(x\right)\right) = \frac{d}{\mathrm{du}} f \left(u\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}} = f ' \left(u\right) \cdot u ' =$

$= f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

There are two ways we could follow:

• We consider $y = {\left[f \left(g \left(x\right)\right)\right]}^{n}$

$f ' \left(x\right) = n {\left[f \left(g \left(x\right)\right)\right]}^{n - 1} \cdot \frac{d}{\mathrm{dx}} g \left(x\right) =$

$= \frac{1}{2} {\left({x}^{2} / \left(3 - x {e}^{x}\right)\right)}^{\frac{1}{2} - 1} \frac{d}{\mathrm{dx}} \left({x}^{2} / \left(3 - x {e}^{x}\right)\right) =$

$= \frac{1}{2} {\left({x}^{2} / \left(3 - x {e}^{x}\right)\right)}^{- \frac{1}{2}} \left(2 x \left(3 - x {e}^{x}\right) - {x}^{2} \frac{0 - \left({e}^{x} + x {e}^{x}\right)}{3 - x {e}^{x}} ^ 2\right) =$

$= \frac{1}{2 \sqrt{{x}^{2} / \left(3 - x {e}^{x}\right)}} \cdot \frac{6 - x {e}^{x} + {x}^{2} {e}^{x}}{{\left(3 - x {e}^{x}\right)}^{2}} \cdot x =$

$= \frac{1}{2} \left(\frac{\sqrt{3 - x {e}^{x}}}{\textcolor{g r e e n}{\cancel{x}}}\right) \frac{6 - x {e}^{x} + {x}^{2} {e}^{x}}{{\left(3 - x {e}^{x}\right)}^{2}} \cdot \textcolor{g r e e n}{\cancel{x}} =$

$= \frac{1}{2} \sqrt{\frac{\textcolor{m a \ge n t a}{\cancel{\left(3 - x {e}^{x}\right)}}}{3 - x {e}^{x}} ^ \left(\textcolor{m a \ge n t a}{{\cancel{4}}^{3}}\right)} \left({x}^{2} {e}^{x} - x {e}^{x} + 6\right) =$

$= \frac{1}{2 \sqrt{{\left(3 - x {e}^{x}\right)}^{3}}} \left({x}^{2} {e}^{x} - x {e}^{x} + 6\right)$

• We consider $y = \sqrt{g \left(x\right)}$

$f ' \left(x\right) = \frac{1}{2 \sqrt{g \left(x\right)}} \frac{d}{\mathrm{dx}} g \left(x\right) =$

$= \frac{1}{2 \sqrt{{x}^{2} / \left(3 - x {e}^{x}\right)}} \cdot \left(\frac{2 x \left(3 - x {e}^{x}\right) - {x}^{2} \left(0 - {e}^{x} + {e}^{x} \left(- x\right)\right)}{3 - x {e}^{x}} ^ 2\right) =$

$= \frac{1}{2} \left(\sqrt{\frac{3 - x {e}^{x}}{x} ^ 2}\right) \cdot \frac{6 x - 2 {x}^{2} {e}^{x} + {x}^{2} {e}^{x} + {x}^{3} {e}^{x}}{{\left(3 - x {e}^{x}\right)}^{2}} =$

$= \frac{1}{2 \textcolor{g r e e n}{\cancel{x}}} \cdot \sqrt{3 - x {e}^{x}} \left(\frac{\textcolor{g r e e n}{\cancel{x}} \left(6 - x {e}^{x} + {x}^{2} {e}^{x}\right)}{{\left(3 - x {e}^{x}\right)}^{2}}\right) =$

$= \frac{1}{2} \sqrt{\frac{\textcolor{m a \ge n t a}{\cancel{\left(3 - x {e}^{x}\right)}}}{3 - x {e}^{x}} ^ \left(\textcolor{m a \ge n t a}{{\cancel{4}}^{3}}\right)} \left({x}^{2} {e}^{x} - x {e}^{x} + 6\right) =$

$= \frac{1}{2 \sqrt{{\left(3 - x {e}^{x}\right)}^{3}}} \left({x}^{2} {e}^{x} - x {e}^{x} + 6\right)$