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# Derivatives - Calculus Tutorial

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!).

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

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

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

(Continue to the tutorials Derivatives 2)

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