# Definition of a Derivative

Our goal in differential calculus is to be able to find a line tangent to any function. This can be achieved using a derivative. A derivative is a special limit that allows us to produce a line tangent to a function. In this section, we will investigate how this is done, and get a good understanding of derivatives.

**Definition: **The tangent line to a curve f(x) at a point (a,f(a)) is the line through P, with the slope $m=lim_{x \to a} \frac{f(x)-f(a)}{x-a}$, such that the limit exists

First, let's get an idea of what this equation is actually telling us. First, it helps to understand that this equation without the limit is essentially the equation for slope, $\frac{y_2-y_1}{x_2-x_1}$. This equation allows us to find slope given two points, however we want to find the slope of a line tangent to a single point. This being the case, we need to adapt the formula to work with a single point.

To do this, we change y_{2} to be the function f(x), and allow it and x_{2} to approach the value a. This will allow us to approach the slope of the single point, without actually equally the single point. This will give us a good approximation for the slope of the tangent. Let's take a look at an example.

**Example: **Find the equation of the tangent line to
the cubic function f(x) = x^{3}, at the point P(2,8)

In this problem, we know that a = 2, so we will place that in our equation and solve it.

$lim_{x \to a} \frac{f(x)-f(a)}{x-a}$

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

$= lim_{x \to 2} \frac{x^3-8}{x-2}$

Once we reach this point, you can see that this is a simple rational function limit. From this point, we can use our limit solving skills from previous lessons to solve the problem.

$= lim_{x \to 2} \frac{x^3-8}{x-2}$

$= lim_{x \to 2} \frac{(x-2)(x^2+2x+4)}{x-2}$

$= lim_{x \to 2} (x^2+2x+4) = 4+4+4 = 12$

This tells us
that the **slope** of the tangent line at the point (2,8) is equal to 12. The
goal of the problem was to find the equation of the tangent line, so we need to
use y = mx+b to find an equation for the line.

$y = mx+b$

$ \Longrightarrow y = 12x+b$

$\Longrightarrow 8 = 12*2+b$

$\Longrightarrow 8 = 24 + b$

$\Longrightarrow b = -16$

Therefore the equation of the tangent line is: $y = 12x-16$

It should be clear how we arrived at this equation. We substituted in the slope we found in the first part of the equation, then substituted in the point (2,8), to solve for the y intercept, b. Following these steps gave us the equation for the tangent line.

There is another way to define the slope of the tangent, that is often used since it is easier to work with. Instead of having x-a at the bottom of the equation, we create a new variable h, such that h = x-a. If we do this, x = a+h, giving us the equation: $lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$. This gives us less variables to worry about, and makes simplifying the limit easier.

To give intuition on why we now have h->0 in the limit, consider what we are trying to do. If h = x-a, then we are trying to get x and a as close as possible. These values are as close as possible when the difference between them is close to 0, therefore we use h->0 in the limit.

Now that we know how to find the slope of the tangent line, we can discuss the derivative in more detail. The derivative of a function is a function that will show the instantaneous rate of change, or slope of the tangent, for any point on the original function. It gives us an equation that will tell us the slope of the tangent without needing to recalculate the limit each time. This is valuable as it greatly simplifies the amount of work required to find a tangent slope.

**Definition: **The derivative of a function, f(x), is denoted as f'(x). This function is defined as $f'(x) = lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

You will notice that this definition is almost the same as the definition of the tangent line. The key difference is that instead of evaluating the function at x=a, we allow it to stay as x, to get a function as a result instead of a single slope. Let's look at the same function as earlier.

**Example: **Find the derivative of the equation f(x)
= x^{3}

Let's go ahead and substitute in f(x) as required, and solve the limit.

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

$=lim_{h \to 0} \frac{(x+h)^3-x^3}{h}$

$=lim_{h \to 0} \frac{x^3+x^2h+2x^2h+2xh^2+xh^2+h^3-x^3}{h}$

$=lim_{h \to 0} \frac{3x^2h+2xh^2+xh^2+h^3}{h}$

$=lim_{h \to 0} \frac{h(3x^2+2xh+xh+h^2)}{h}$

$=lim_{h \to 0} 3x^2+2xh+xh+h^2$

$=3x^2$

Therefore, $f'(x) = 3x^2$. We can verify this function by substituting in the point from our previous problem and see if the slopes match up. At (2,8), we get: $f'(2) = 3*(2*2) = 12$, which matches the previous problem.

There are different notations used for derivatives, so you may see $f'(x)$ or you may see $\frac{dy}{dx}$. Both notations mean the same thing.

Based on the definition of a derivative, we can conclude that a derivative exists where ever the limit is defined. A function can fail to be differentiable in three main conditions:

- There is a sharp corner
- There is a discontinuity
- A vertical tangent

Aside from these three conditions, we are able to use derivatives to determine the rate of change of a function at any point.