Calculus Tutorials - Derivatives (II)
On the second part of this tutorial, we'll work on some other
slightly more complicated examples.
Example: Given the function  = x^3 + 2x+1) , compute
the derivative ) 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  . How do we do that?
Well, we just follow the definition:
 = \lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0})
and now we use the definition of ) . In fact, we obtain:
 = \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} )
Now we use a little and neat algebraic trick:
(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
 = \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  , and we finally get
 = \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  = 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
 .
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.
|
|
