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
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 $\displaystyle f(x) = +\sqrt {x^2}$ for positive $\displaystyle x$ and then do the same but minus it for negative $\displaystyle x$. You'll get two expressions that you can combine by using the abs function as suggested in the answer you're directed towards.
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 $\displaystyle \frac {-x} x$ 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 $\displaystyle d/dx(abs x)$.
So you use the identity that $\displaystyle abs x = +\sqrt {x^2}$.
So you differentiate $\displaystyle \sqrt {x^2}$ using the chain rule, noticing that it's $\displaystyle 1$ for $\displaystyle x > 0$ and it's $\displaystyle -1$ for $\displaystyle x < 0$.
So, oh good, we know the driv of $\displaystyle \sqrt {x^2}$.
But this is the same as the driv of $\displaystyle abs x$.
So $\displaystyle d/dx(abs x)$ equals 1 for x > 0 and -1 for x < 0.
That's the main result.
The final bit is showing that $\displaystyle abs x / x$ 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.