1. Chain Rule

Write absolute value(x)=sqrt(x^2) and use the chain rule to show that

(d/dx)abs(x)=x/abs(x)

could i get some help with this please?

i know the chain rule is F'(x)=f'(g(x))*g'(x) but i do not know how to apply that to this problem.

thank you

2. First you need to divide the domain into two bits to get rid of that pesky "absolute value" bit.

When x > 0, abs x = x.
When x < 0, abs x = -x.

So first differentiate $f(x) = +\sqrt {x^2}$ for positive $x$ and then do the same but minus it for negative $x$. You'll get two expressions that you can combine by using the abs function as suggested in the answer you're directed towards.

3. i kinda get what you are saying but i guess im just kind of confused because isnt the sqrt(x^2) just x? and the derivative of x is just 1?

4. Originally Posted by mybrohshi5
i kinda get what you are saying but i guess im just kind of confused because isnt the sqrt(x^2) just x? and the derivative of x is just 1?
Yep, that's right. Except that the derivative for x < 1 is -1 (you have to take the negative of it, remember?).

So: what's abs x when x > 1? It's x.

So what's abs x / x when x > 1? It's x / x = 1.

Now: what's abs x when x < 1? It's -x.

So what's abs x / x when x < 1? It's $\frac {-x} x$ which is .... ?

5. it is -1,
but i dont get how this is helping to show that (d/dx)abs(x)=x/abs(x)

i mean i understand what you are doing for the most part but im not getting how to put this all together to help show that (d/dx)abs(x)=x/abs(x)

thank you very much =)

6. You want to find $d/dx(abs x)$.

So you use the identity that $abs x = +\sqrt {x^2}$.

So you differentiate $\sqrt {x^2}$ using the chain rule, noticing that it's $1$ for $x > 0$ and it's $-1$ for $x < 0$.

So, oh good, we know the driv of $\sqrt {x^2}$.

But this is the same as the driv of $abs x$.

So $d/dx(abs x)$ equals 1 for x > 0 and -1 for x < 0.

That's the main result.

The final bit is showing that $abs x / x$ is *also* equal to 1 when x > 0 and -1 when x < 0.

Equate the two.

Bingo. You've now proved what you were asked to.

7. If you still have problems, get someone else to help because it's nearly 1:30 and I'm going off line.

8. is it because the f'(x) of +sqrt(x^2) is 1 and f'(x) of -sqrt(x^2) is -1 so....

(d/dx)abs(x) for x>1 will be +1 and for (d/dx)abs(x) for x<1 will be -1?

9. thank you =) i get it all now