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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

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

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

(Continue to the tutorials Derivatives 2)

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