# Quotient Rule

## Key Questions

• $y ' = \frac{g \left(x\right) f ' \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$

$y = f \frac{x}{g} \left(x\right) = \frac{2 {x}^{4} - 3 x}{4 x - 1}$

$f ' \left(x\right) = 8 {x}^{3} - 3$

$g ' \left(x\right) = 4$

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

$y ' = \frac{\left(4 x - 1\right) \left(8 {x}^{3} - 3\right) - \left(2 {x}^{4} - 3 x\right) \left(4\right)}{4 x - 1} ^ 2$

$y ' = \frac{32 {x}^{4} - 12 x - 8 {x}^{3} + 3 - 8 {x}^{4} + 12 x}{4 x - 1} ^ 2$

Simplify for combining like terms.

$S o l u t i o n \to y ' = \frac{24 {x}^{4} - 8 {x}^{3} + 3}{4 x - 1} ^ 2$

• $y ' = \frac{1}{\sqrt{x}} \cdot \frac{1}{{\left(1 - \sqrt{x}\right)}^{2}}$

Explanation :

Using Quotient Rule, which is

$y = f \frac{x}{g} \left(x\right)$, then

$y ' = \frac{g \left(x\right) f ' \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$

Similarly following for the given problem,

$y = \frac{1 + \sqrt{x}}{1 - \sqrt{x}}$

$y ' = \frac{\left(1 - \sqrt{x}\right) \left(\frac{1}{2 \sqrt{x}}\right) - \left(1 + \sqrt{x}\right) \left(- \frac{1}{2 \sqrt{x}}\right)}{{\left(1 - \sqrt{x}\right)}^{2}}$

$y ' = \frac{1}{2 \sqrt{x}} \cdot \frac{1 - \sqrt{x} + 1 + \sqrt{x}}{{\left(1 - \sqrt{x}\right)}^{2}}$

$y ' = \frac{1}{2 \sqrt{x}} \cdot \frac{2}{{\left(1 - \sqrt{x}\right)}^{2}}$

$y ' = \frac{1}{\sqrt{x}} \cdot \frac{1}{{\left(1 - \sqrt{x}\right)}^{2}}$

((u(x))/(v(x)))^'=(u^'(x)*v(x)-u(x)*v'(x))/((v(x))²)

#### Explanation:

Assuming that those who are reading have a minimum level in Maths, everyone knows perfectly that the quotient rule is color(blue)(((u(x))/(v(x)))^'=(u^'(x)*v(x)-u(x)*v'(x))/((v(x))²)), where $u \left(x\right)$ and $v \left(x\right)$ are functions and $u ' \left(x\right)$, $v ' \left(x\right)$ respective derivates. But, where does it come from?
Let's find out!

Considering $f \left(x\right) = \frac{u \left(x\right)}{v \left(x\right)}$, and by definition, $f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$
So :
$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\frac{u \left(x + h\right)}{v \left(x + h\right)} - \frac{u \left(x\right)}{v \left(x\right)}}{h}$

Now we need a common denominator :

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\frac{u \left(x + h\right) \textcolor{red}{\cdot v \left(x\right)}}{v \left(x + h\right) \textcolor{red}{\cdot v \left(x\right)}} - \frac{u \left(x\right) \textcolor{red}{\cdot v \left(x + h\right)}}{v \left(x\right) \textcolor{red}{\cdot v \left(x + h\right)}}}{h}$
$f ' \left(x\right) = {\lim}_{h \to 0} \frac{u \left(x + h\right) \textcolor{red}{\cdot v \left(x\right)} - u \left(x\right) \textcolor{red}{\cdot v \left(x + h\right)}}{v \left(x\right) \cdot v \left(x + h\right)} \cdot \frac{1}{h}$
$f ' \left(x\right) = {\lim}_{h \to 0} \frac{u \left(x + h\right) \textcolor{red}{\cdot v \left(x\right)} - u \left(x\right) \textcolor{red}{\cdot v \left(x + h\right)}}{h \cdot v \left(x\right) \cdot v \left(x + h\right)}$

This expression isn't very useful for the moment, so let's add an intelligent 0:

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{u \left(x + h\right) v \left(x\right) - u \left(x\right) v \left(x + h\right) \textcolor{red}{+ u \left(x\right) v \left(x\right) - u \left(x\right) v \left(x\right)}}{h v \left(x\right) v \left(x + h\right)}$
Now we can factoring :

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{v \left(x\right) \left(u \left(x + h\right) - u \left(x\right)\right) + u \left(x\right) \left(v \left(x\right) - v \left(x + h\right)\right)}{h v \left(x\right) v \left(x + h\right)}$
Now we can cut our limit into two limits :
$f ' \left(x\right) = {\lim}_{h \to 0} \frac{v \left(x\right) \left(u \left(x + h\right) - u \left(x\right)\right)}{h v \left(x\right) v \left(x + h\right)} + {\lim}_{h \to 0} \frac{u \left(x\right) \left(v \left(x\right) - v \left(x + h\right)\right)}{h v \left(x\right) v \left(x + h\right)}$
$f ' \left(x\right) = {\lim}_{h \to 0} \frac{v \left(x\right)}{v \left(x\right) v \left(x + h\right)} \cdot {\cancel{\frac{u \left(x + h\right) - u \left(x\right)}{h}}}^{= u ' \left(x\right)} - {\lim}_{h \to 0} \frac{u \left(x\right) \left(v \left(x + h\right) - v \left(x\right)\right)}{h v \left(x\right) v \left(x + h\right)}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{v \left(x\right) u ' \left(x\right)}{v \left(x\right) \left(v \left(x + h\right)\right)} - {\lim}_{h \to 0} \frac{u \left(x\right)}{v \left(x\right) v \left(x + h\right)} \cdot {\cancel{\frac{v \left(x + h\right) - v \left(x\right)}{h}}}^{= v ' \left(x\right)}$
$f ' \left(x\right) = {\lim}_{h \to 0} \frac{v \left(x\right) u ' \left(x\right)}{v \left(x\right) \left(v \left(x + h\right)\right)} - {\lim}_{h \to 0} \frac{u \left(x\right) v ' \left(x\right)}{v \left(x\right) v \left(x + h\right)}$
$f ' \left(x\right) = {\lim}_{h \to 0} \frac{v \left(x\right) u ' \left(x\right) - u \left(x\right) v ' \left(x\right)}{v \left(x\right) v \left(x + h\right)}$
And because v(x+h)≈_(h to 0)v(x),
f'(x)=(u^'(x)*v(x)-u(x)*v'(x))/((v(x))²)