## Calculus Tutorial

#### Intro

Any problem that can be solved with the quotient rule can also be solved with the product rule and chain rule.

#### Sample Problem

Find the first derivative of:

$f(x)=\frac{cos(2x)}{x^3 - 2x}$

#### Solution

Quotient rule:

Separate the numerator and denominator into two separate functions, then find their derivatives.

$g(x)=cos(2x)$
$h(x)=x^3&space;-2x$

$g'(x)=-2sin(2x)$
$h'(x)=3x^2&space;-2$

Plug into the quotient rule formula:

$f(x) = \frac{g(x)}{h(x)} =\frac{g'(x)h(x) - g(x)h'(x)}{g^{2}(x)}$

$f'(x)=\frac{[-2sin(2x)(x^3 - 2x)] - [cos(2x)(3x^2 - 2)]}{(x^3 - 2x)^2}$

You can expand the answer if you like, but this is sufficient.

————————————————————-

Now for the Product Rule. This time, we will turn the denominator into a negative exponent:

$f(x)&space;=&space;cos(2x)[x^3&space;-&space;2x]^-^1$

g(x) will remain the same, but h(x) and h'(x) will look like this using the chain rule:

$h(x)&space;=&space;(x^3&space;-&space;2x)^-^1$

$h'(x)&space;=&space;-(x^3&space;-&space;2x)^-^2\cdot&space;[3x^2-2]$

And plug into the product rule:

$f'(x)&space;=&space;g(x)h(x)&space;=&space;g'(x)h(x)&space;+&space;g(x)h'(x)$

$f'(x)&space;=&space;-2sin(2x)[x^3&space;-&space;2x]^-^1&space;+&space;cos(2x)[-(x^3-2x)^-^2&space;\cdot&space;(3x^2-2)]$

$f'(x)&space;=&space;\frac{-2sin(2x)}{x^3&space;-&space;2x}&space;-&space;\frac{cos(2x)\cdot&space;(3x^2-2)}{(x^3-2x)^2}$

$f'(x)&space;=&space;\frac{-2sin(2x)}{x^3&space;-&space;2x}&space;\cdot&space;\frac{x^3&space;-&space;2x}{x^3&space;-&space;2x}-&space;\frac{cos(2x)\cdot&space;(3x^2-2)}{(x^3-2x)^2}$

$f'(x)=\frac{[-2sin(2x)(x^3 - 2x)] - [cos(2x)(3x^2 - 2)]}{(x^3 - 2x)^2}$

You should arrive at the same answer in both cases

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