# Quotient Rule

How to differentiate tanx using the quotient rule

Tip: This isn't the place to ask a question because the teacher can't reply.

1 of 2 videos by Tiago Hands

## 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,

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

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

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

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

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

• This key question hasn't been answered yet.

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