# The Basic Concept of Derivatives

Imagine you have a function $$f(x)$$. For example you could have something like $$f(x) = x^2$$ or maybe something like $$f(x) = \sin x$$. We define the derivative of the function $$f(x)$$ at the point $$x_0$$ as

$f'(x_0) = \lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}$

if the limit exists. Before you complain saying "What the heck is this??" let me tell you something, this is not complicated as it may look at first sight. First, a couple of observations about what is this limit all about.

• The derivative $$f'(x)$$ is also a function (whenever it is defined).

• The derivative is computed at a given point $$x_0$$, using the limit shown above. If this limit exists, and only if it exists, we say that the derivative is well defined at the point $$x_0$$ a, and it is written as $$f'(x_0)$$

• In other words, the derivative $$f'(x)$$ can be thought as a function that depends on the original function $$f(x)$$, and which is computed point by point.

• That's it, that's all you need to know for now (seriously!).

Observe that the process of calculating derivatives is not too complicated, but life can be made easier by this derivative calculator, which shows you all the steps of the process.

## The Derivative as a Function

Observe that the concept of derivative at a given point $$x_0$$ is interpreted as the instant rate of change of the function at that point. This is achieved by computing the average rate of change for an interval of width $$\Delta x$$, and taking that $$\Delta x$$ as it approaches to zero.

It is time to go for some neat examples to understand what's going on.

## Derivative Examples

Example : Compute the derivative of the function $$f(x) = x^2$$ at the point $$x_0 = 2$$

Solution : We simply use the definition and replace the corresponding terms. Let's see what we get:

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

We simply replaced $$f(x) = x^2$$ and $$x_0 = 2$$ in the original definition of derivative. Now, noticing that $$x^2 - 2^2 = (x-2)(x+2)$$, we find that

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

In the next tutorial we'll learn more things about how to compute derivatives.

(Continue to the tutorials Derivatives 2 )