Expert: Scotto - 5/6/2010

Question
QUESTION: Hey, im in my last year of high school, but im doing Uni Maths (basic, id guess), and i dont quite get it all...

how does one differentiate f(x)=x(1+|x|) ?

i dont quite understand all the limit stuff, and how it works with the modulus etc.

any thoughts?

ANSWER: Here, f(x) is really two functions.
For x>=0, f(x) = x(1+x) = x + x².
FOr x<=0, f(x) = x(1-x) = x - x².

This done, we can now find the derivative.
For x>0, f'(x) = 1 + 2x.
For x<0, f'(x) = 1 - 2x.

In order for f'(0) to exist, the values of both would have to be the same.
Putting 0 into the 1st gives 1 + 2(0) = 1 and putting 0 into the 2nd gives 1 - 2(0) = 1.
Thus, f'(0) is defined as well by both functions.  We could then say
For x>=0, f'(x) = 1 + 2x and for x<=0, f'(x) = 1 - 2x since the value at 0 is 1 for both cases.


---------- FOLLOW-UP ----------

QUESTION: yep, i get that, but how would you prove that f'(0)=1 using limits?

Answer
A proof of a formula on limits based on the epsilon-delta definition.
Here, the function is looked at from the left and the right.

The theorem on derivatives is an epsilon-delta proof.
It says that no matter how small of a value of epsiln we choose,
the derivative must be within ±delta, where delta is a function of epsilon.

Let us take the difference as 0.1.  The derivative is f'(x) = 1 + 2x, so f'(0.1) = 1.2.
Let us take delta as 20*epsilon { 20 is big enough; it could also be 40, 82, 52.3;
but I can see that 20 is good enough }.

Note that f(0.1) = 0.1*(1+0.1) = 0.11 and that f(0) = 0(1+0) = 0.
So at 0.1, we have (f(0.1) - f(0))/(0.1 - 0) = (0.11 - 0)/0.10 = 11/10 = 1.1.

Using 0.1, the theory says if we multiply by 20, we get a bigger number.
Since 0.1*20 = 2, and 2 is greater than 1.1, this is true.

Maybe if we took -0.0001.  Now would that would be too close?
Well, it is known that f(-0.0001) = -0.0001(1.0001) = -0.00010001.  The abolute value of this is 0.00010001.  Now for delta, we get 20*0.0001 = 0.0020, and that can be seen to be greater than 0.00010001.

So now matter how close we get with epsion, we have the formula for delta to make it closer.



Another way to look at it is as f(x) = x + x|x|.
The x has slope 1, no matter where we are.
The x|x| is the same as x² (if x is positive) or -x² (if x is negatve).
Both of the functions, when x is taken to 0, approach 0, so the derivative from x|x| is 0 at x=0.
Hence the only part to worry about is f(x) = x, for x|x| has been shown to have 0 slope at x=0.
The derivative of x is 1.