11-24-2011, 02:45 PM

Lesson 2 - Intro to Derivatives

The second important concept is that of the derivative. Before calculus, we are mostly focused on finding the slope of a line, or the rate of change. In calculus, however, we are more concerned with instantaneous rate of change, or the slope of a curve. Therefore, the derivative at a point is also the slope of the tangent line at that point. With this concept we can graph most equations and lead into more complex concepts.

Finding the Derivative

The simplest way to find the derivative is using a simple equation.

Code:

`f′(x) = n*x^(n-1)`

So, the derivative of:

Code:

`f(x) = 2x^3 is`

f′(x) = 6x^2

It is important to note that the derivative of a constant is 0.

Using the Derivative

It may seem like the concept is not that helpful, but the truth is the opposite. With the derivative we can find several things.

Let's say we have the equation.

Code:

`x^3 + x^2 + 4x + 1`

How many minima and maxima does the equation have? To figure this out we would find the first derivative. There is also the question of decreasing or increasing. If f′(x) is negative, the graph is decreasing. If it is positive, it is increasing.

Code:

`f′(x) = 3x^2 + 2x + 4`

We could then find the zeros of the first derivative, giving us the points where the original equation turns from decreasing to increasing and vice-versa.

Code:

`f′(x) = (3x + 1)(x + 1)`

f′(0) = 1/3 and f′(0) = -1

So, the graph has an extrema (min or max) at x = 1/3 and x = -1.

What about turning points and concavity? For that we need a higher derivative. If the second derivative (f′′(x)) is negative, the graph is concave down. If it is positive it is concave up. Points of inflection are defined where the second derivative is 0.

Code:

`f′′(x) = 6x + 2`