1. ## derivative

Im supposed to find the derivative of the following two problems:

(a) f(x) = |(x+2)/(x-3)|

(b) f(x) = (1/(1+|x|))+(1/(1+|x-3|))

I have to mention my teacher just havent bother to teach how to derive of the absolute value... so if any one can also explain that to me greatly apreciated...

also if anything sounds weird im sorry but im talking math clases in spanish for the first time in long time.

thanx

2. Originally Posted by sacwchiri
I have to mention my teacher just havent bother to teach how to derive of the absolute value... so if any one can also explain that to me greatly apreciated...
The derivative of the function f=|x| is:
1 if x>0
-1 if x<0

Note it is not differencialable at x=0.

Another way to write the absolute value derivative is,
x/|x|

3. Originally Posted by sacwchiri
Im supposed to find the derivative of the following two problems:

(a) f(x) = |(x+2)/(x-3)|

(b) f(x) = (1/(1+|x|))+(1/(1+|x-3|))

I have to mention my teacher just havent bother to teach how to derive of the absolute value... so if any one can also explain that to me greatly apreciated...

also if anything sounds weird im sorry but im talking math clases in spanish for the first time in long time.

thanx
Consider y = |x|. This function is: {y = -x for x < 0 and y = x for 0<= x
} This means the derivative of y is {y' = -1 for x < 0 and y' = 1 for 0 < x}. Note that the derivative of y is undefined for x = 0.

My best advice is to write the function out in terms of its domains before you take the derivative. For example:

f(x) = |(x+2)/(x-3)|

Use the intervals: (-infinity, -2), (-2, 3), and (3, infinity).
On (-infinity, -2) f(x) = -(x+2)/[-(x-3)] = (x+2)/(x+3)
On (-2, 3) f(x) = (x+2)/[-(x-3)] = -(x+2)/(x-3)
On (3, infinity) f(x) = (x+2)/(x-3)

Now take the derivative of each of the expressions above. Again we have that the derivative of f(x) does not exist for x = -2, nor for x = 3.

-Dan

4. Consider a differenciable function f(x) on the interval (a,b)

What is the derivative of |f(x)|?

To answer the question we use the chain rule,
Thus,
f'(x) if f(x)>0
-f'(x) if f(x)<0

And the zeros of f(x) on (a,b) do not make |f(x)| differenciable.

5. ## derivative

thanx... this has been really helpfull