notes/education/math/MATH1210 (calc 1)/Derivatives.md

A derivative can be used to describe the rate of change at a single point, or the *instantaneous velocity*.

The formula used to calculate the average rate of change looks like this:
$$ \dfrac{f(b) - f(a)}{b - a} $$
Interpreting it, this can be described as the change in $y$ over the change in $x$.

- Speed is always positive
- Velocity is directional

As the distance between the two points $a$ and $b$ grow smaller, we get closer and closer to the instantaneous velocity of a point. Limits are suited to describing the behavior of a function as it approaches a point.

If we have the coordinate pair $(a, f(a))$, and the value $h$ is the distance between $a$ and another $x$ value, the coordinates of that point can be described as ($(a + h, f(a + h))$. With this info:
- The slope of the secant line can be described as $\dfrac{f(a + h) - f(a)}{a + h - a}$, which simplifies to $\dfrac{f(a + h) - f(a)}{h}$. 
- The slope of the *tangent line* or the *instantaneous velocity* can be found by taking the limit of the above function as the distance ($h$) approaches zero:
$$\lim_{h \to 0}\dfrac{f(a + h) - f(a)}{h}$$
The above formula can be used to find the *derivative*. This may also be referred to as the *instantaneous velocity*, or the *instantaneous rate of change*.
# Line Types
## Secant Line
A **Secant Line** connects two points on a graph.

A **Tangent Line** represents the rate of change or slope at a single point on the graph.

# Notation
Given the equation $y = f(x)$, the following are all notations used to represent the derivative of $f$ at $x$:
- $f'(x)$
- $\dfrac{d}{dx}f(x)$
- $y'$
- $\dfrac{dy}{dx}$
- $\dfrac{df}{dx}$
- "Derivative of $f$ with respect to $x$"

# Functions that are not differentiable at a given point
- Where a function is not defined
- Where a sharp turn takes place
- If the slope of the tangent line is vertical

# Higher Order Derivatives
- Take the derivative of a derivative

# Exponential Derivative Formula
Using the definition of a derivative to determine the derivative of $f(x) = x^n$, where $n$ is any natural number.

$$ f'(x) = \lim_{h \to 0} \dfrac{(x + h)^n - x^n}{h} $$
- Using pascal's triangle, we can approximate $(x + h)^n$
```
    1
   1 1
  1 2 1
 1 3 3 1
1 4 6 4 1 
```

- Where $n = 0$: $(x + h)^0  = 1$
- Where $n = 1$: $(x +h)^1 = 1x + 1h$
- Where $n = 2$: $(x +h)^2 = x^2 + 2xh + h^2$
- Where $n = 3$: $(x + h)^3 = 1x^3h^0 + 3x^2h^1 + 3x^1h^2 + 1x^0h^3 = 1x^3 + 3x^2h + 3xh^2 + 1h^3$

Note that the coefficient follows the associated level of Pascal's Triangle (`1 3 3 1`), and $x$'s power decrements, while $h$'s power increments. The coefficients of each pair will always add up to $n$. Eg, $3 + 0$, $2 + 1$, $1 + 2$, and so on. The **second** term in the polynomial created will have a coefficient of $n$. 

$$ \dfrac{(x + h)^n - x^n}{h} = \lim_{h \to 0} \dfrac{(x^n + nx^{n-1}h + P_{n3}x^{n-2}h^2 + \cdots + h^n)-x^n}{h} $$ $P$ denotes some coefficient found using Pascal's triangle.

$x^n$ cancels out, and then $h$ can be factored out of the binomial series. 

This leaves us with:
$$ \lim_{h \to 0} nx^{n-1} + P_{n3} x^{n-2}*0 \cdots v * 0 $$

The zeros leave us with:

$$ f(x) = n, \space f'(x) = nx^{n-1} $$
# Sum and Difference Rules
$$ \dfrac{d}{dx}(f(x) \pm g(x)) = f'(x) \pm g'(x) $$

# Product Rule
$$ \dfrac{d}{dx} (f(x) * g(x)) = \lim_{h \to 0} \dfrac{f(x +h) * g(x + h) - f(x)g(x)}{h} $$
This is done by adding a value equivalent to zero to the numerator ($f(x + h)g(x) - f(x + h)g(x)$):
$$ \dfrac{d}{dx} (f(x) * g(x)) = \lim_{h \to 0} \dfrac{f(x +h) * g(x + h) + f(x + h)g(x) - f(x+h)g(x) - f(x)g(x)}{h} $$

From here you can factor out $f(x + h)$ from the first two terms, and a $g(x)$ from the next two terms.

Then break into two different fractions:

$$\lim_{h \to 0} \dfrac{f(x + h)}{1} * \dfrac{(g(x + h) - g(x))}{h)} + \dfrac{g(x)}{1} *\dfrac{f(x + h) - f(x)}{h} $$
From here, you can take the limit of each fraction, therefore showing that to find the derivative of two values multiplied together, you can use the formula:
$$ \dfrac{d}{dx}(f(x) * g(x)) = f(x) * g'(x) + f'(x)*g(x) $$

# Constant Multiple Rule
$$ \dfrac{d}{dx}[c*f(x)] = c * f'(x) $$
# Quotient Rule
$$ \dfrac{d}{dx}(\dfrac{f(x)}{g(x)}) = \dfrac{f'(x)g(x) -f(x)g'(x)}{(g(x))^2}  $$

# Exponential Rule
$$ \dfrac{d}{dx} e^x = e^x $$
$$ \dfrac{d}{dx}a^x = a^x*(\ln(a)) $$
for all $a > 0$