Write absolute value(x)=sqrt(x^2) and use the chain rule to show that
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.
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 for positive and then do the same but minus it for negative . You'll get two expressions that you can combine by using the abs function as suggested in the answer you're directed towards.
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 which is .... ?
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)
could you please explain
thank you very much =)
You want to find .
So you use the identity that .
So you differentiate using the chain rule, noticing that it's for and it's for .
So, oh good, we know the driv of .
But this is the same as the driv of .
So equals 1 for x > 0 and -1 for x < 0.
That's the main result.
The final bit is showing that 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.