Question

Can anyone help me please?
http://i.imgur.com/ai1qawa.jpg
I don't understand absolute values when it comes to derivatives...

Answer 1

For absolute value, in order to take the derivative tou have to rewrite it like this:

\[|x| = \sqrt{x^{2}}\]When you have it like that you can take the derivative like normal.

Answer 2

what you get your math questions from a fortune cookie?

Answer 3

lolol

Answer 4

no,from my calc textbook lol.

Answer 5

Maybe your textbook is from panda express o.o

Answer 6

@Psymon do i need to add the plus or minus sign in that form?

Answer 7

Nah, not really. Youll never really be taking it out of the absolute value bars. At the end, youll have x/sqrt(x^2), which you can just simplify back into terms of absolute value as x/|x|

Answer 8

i would go with
\[ |x| = \left\{\begin{array}{rcc}
-x & \text{if} & x <0 \\
x& \text{if} & x >0
\end{array}
\right.\]
and work in cases although @Psymon method will also work

Answer 9

but i think that might be easier because you get the derivative right away
you have
\[ x|x| = \left\{\begin{array}{rcc}
-x^2 & \text{if} & x <0 \\
x^2& \text{if} & x >0
\end{array}
\right.\]

Answer 10

Right. If you had to evaluate anythign then you definitely want to be aware of the above definition of absolute value. But I think its fine to not worry about it too much if its just a derivative?

is it preferred to put the derivatives of absolute values as piecewises like that? Ive seen people do it but never thought of it, lol.

Answer 11

then the derivative is
\[f'(x) = \left\{\begin{array}{rcc}
-2x & \text{if} & x <0 \\
2x& \text{if} & x >0
\end{array}
\right.\]

Answer 12

which exits at \(x=0\) as they match up there, but \[f''(x) = \left\{\begin{array}{rcc}-2& \text{if} & x <0 \\
2& \text{if} & x >0
\end{array}
\right.\]

Answer 13

yeah i think peicewise is nice because you get something real easy to work with in this case, namely \(x^2\) and \(-x^2\) but i guess it makes no real difference

Answer 14

it is also easy now to see that the derivative does not exist at \(x=0\) because \(-2\neq 2\)

Answer 15

thanks guys. :> im going to reread this until i finally get it 0-0

Answer 16

i meant "the second derivative does not exist at \(x=0\)" sorry