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 Differentials

• Take the differential of a differential

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 = 1