Derivative/Definition

The derivative of a function is defined as the instantaneous rate of change of the function at a certain point. For a line, this is just the slope. For more complex curves, we can find the rate of change between two points on the curve easily since we can draw a line through them.

In the image above, the average rate of change between the two points is the slope of the line that goes through them: $\frac{f(x+h)-f(x)}h$ .

We can move the second point closer to the first one to find a more accurate value of the derivative. Thus, taking the limit as $h$ goes to 0 will give us the derivative of the function at $x$ :

$f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}h.$

If this limit exists, it is the derivative of $f$ at $x$ . If it does not exist, we say that $f$ is not differentiable at $x$ . This limit is called Fermat's difference quotient.

Examples

We can apply the Fermat's difference quotient to a polynomial of the form $f(x)=ax^n+bx^{n-1}+cx^{n-2}+ \cdots + z=0$ in order to find its derivative. If we imagine the secant line intersecting a curve at the points $A$ and $B$ . Then we can change this to the tangent by setting $B$ on top of $A$ . Let us call the horizontal or vertical distance as $h$ .

$\lim_{h\to0} \frac{f(x+h)-f(x)}{h}$

$\implies \lim_{h\to0} \frac{a(x+h)^n+b(x+h)^{n-1}+c(x+h)^{n-2}+ \cdots z-(ax^n+bx^{n-1}+cx^{n-2}+ \cdots z)}{h}$

After canceling like terms we should have all terms contain an $h$ . We can then cancel out the $h$ and set $h=0$ . Our end result is the first-derivative.

The first derivative is denoted as $f'(x)$ .

This would be some tedious work so instead there is a much nicer way to find the derivative.

Let $f(x)=3x^n$ . Let $g(x)=x^t+x^{n-1}+5x^{3}$

1. Find $f'(x)$ .

Any function like this is: $f'(x)=3 \cdot n \cdot x^{n-1}=3n \cdot x^{n-1}$

2. Find $g'(x)$ .

Breaking apart on what we used above.

$g'(x)=t \cdot x^{t-1}+(n-1) \cdot x^{n-2}+ 5 \cdot 3 \cdot x^2$

$g'(x)=t \cdot x^{t-1}+(n-1) \cdot x^{n-2}+15x^2$

Let $f(x)=-147$ . Find $f'(x)$ .

If the function $f(x)$ is a constant then its derivative will always be $0$ .

Notation: $f'(x)$ denotes the first derivative for $f(x)$ . The symbol for the second derivative is just $f''(x)$ . For the third derivative it is just $f'''(x)$ . Derivatives are also written as $\frac{d}{dx} f(x)$ . Or if for the nth derivative they are written as $\frac{d^n}{dx^n} f(x)$ .

Maximum and Minimum: We can use the first derivative to determine the maximum and the minimum points of a graph.

If $f'(x)=6x^2-24$ . Then the maximum and the minimum occur when:

$6x^2-24=$ , $x=2$ or $x=-2$ . We can plug each back in to the original $f(x)$ if it was given, and the one with the higher y-coordinate is the maximum, while the smaller y-coordinate gives the minimum.

Below are problems for Part I. In Part II(see link below) we will begin to actually "start" the calculus with this.

Problems

$\boxed{\text{Problem 1}}$ : Find the first derivative of $f(x)$ , where $f(x)=2x^2-15x+7$ .

$\boxed{\text{Solution 1}}$ :

$f'(x)=2 \cdot 2 \cdot x^1-15 \cdot 1 \cdot x^0+0$

$f'(x)=4x-15$ .

$\boxed{\text{Problem 2}}$ : Find the equation of the line tangent to the function $f(x)=3x^3-5x^2+12$ at the point $(-1, 4)$ .

$\boxed{\text{Solution 2}}$ :

We will take the first derivative to determine the slope of the tangent line.

$f'(x)=9x^2-10x$ . If this is the slope of the tangent point then we can just plug $-1$ into the $x$ coordinate to find the actual slope.

$f'(x)=9+10=19$ . The slope of the line is $19$ .

Let the equation be:

$y=19x+b$ .

Plugging $(-1,4)$ in gives $4=-19+b$ and so $b=23$ .

Thus, the equation of the line is $y=19x+23$ . Alternatively, one could use point-slope form for the line; after determining that the slope is $19$ , as above, this allows one to immediately write down the equation $y - 4 = 19(x + 1)$ of the line.

(Notice that it is implicit in the question that the point $(-1, 4)$ lies on the graph of $y = f(x)$ ; it's easy to check that this is actually the case.)