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# Derivatives (II) - Calculus Tutorials

On the second part of this tutorial, we'll work on some other slightly more complicated examples.

Example: Given the function $$f(x) = x^3 + 2x+1$$, compute the derivative $$f'(x)$$ for every every point where it is defined.

Solution: Notice that in this problem they are not giving us an specific point at which to compute the derivative. We need to compute at an arbitrary point $$x_0$$. How do we do that? Well, we just follow the definition:

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

and now we use the definition of $$f(x)$$. In fact, we obtain:

$f'(x_0) = \lim_{x\to x_0} \frac{(x^3+2x+1)-(x_0^3+2x_0 +1)}{x-x_0} =\lim_{x\to 0} \frac{x^3+2x+1-x_0^3-2x_0 -1}{x-x_0}$ $= \lim_{x\to x_0} \frac{x^3-x_0^3 + 2x-2x_0}{x-x_0}$

Now we use a little and neat algebraic trick:

$x^3 - x_0^3 = (x-x_0)(x^2+xx_0+x_0^2)$

Now pay attention. We use this little trick in the last part of the calculation of the derivative, and we find that

$f'(x_0) = \lim_{x\to x_0} \frac{x^3-x_0^3 + 2x-2x_0}{x-x_0} = \lim_{x\to 0} \frac{(x-x_0)(x^2+xx_0+x_0^2) + 2x-2x_0}{x-x_0}$

As you can observe, we can cancel $$x-x_0$$, and we finally get

$f'(x_0) = \lim_{x\to x_0} (x^2+xx_0+x_0^2 + 2) = x_0^2 + x_0^2+x_0^2+2 = 3x_0^2 +2$

In other words, the derivative function is $$f'(x) = 3x^2+2$$. You see? That's what I meant when I said that the derivative is also a function. In this case, the derivative is well defined for all $$x\in \mathbb R$$.

Yes, it is true that we needed some tricks to compute the derivative. So, how are you going to do it?? Let me tell you something, you won't be computing derivatives by hand like that most of the time. On the next tutorial, I'll introduce you with some tools that make very easy the calculation of derivatives.

So, hang on until the next one.

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