Question

weird but no one taught me in school how to take a derivative of an absolute value... tell me is I am doing it correctly.

Answer 1

I am taking the derivative of\[\left| x^3 \right|\]
re-writing,
\[\sqrt{(x^3)^2}\]
I will post this so far, because the post button is hiding

Answer 2

I mean the equation editor is hiding, anyway...

Answer 3

You could always write it as a piecewise function and differentiate
OR
use the chain rule and use the following:
\[(|x|)'=\frac{x}{|x|}\]

Answer 4

\[\frac{x}{dx}~\sqrt{(x^3)^2}~=~\frac{1}{2\sqrt{(x^3)^2}} \times 6x^5\]so, I would be getting:
\[\frac{x}{dx}~\sqrt{(x^3)^2}~= \frac{3x^5}{\left| x^3 \right|}\]

Answer 5

am I differentiating correctly?

Answer 6

if so, then also, I should be able to take out the x^2, or not?

Answer 7

Looks good.
\[(|x^3|)'=\frac{x^3}{|x^3|}(x^3)'=\frac{x^3}{|x^3|}3x^2=\frac{3x^5}{|x^3|}\]

Answer 8

*

Answer 9

a bit complicated :|
when u deal with absolute value remember this
REWRITE IT AS PIECEWISE

Answer 10

can I make it \[\frac{3x^5}{\left| x^3 \right|}=\frac{3x^3}{\left| x\right|}\] ?

Answer 11

x^2>0
so |x^3|=|x^2*x|=|x^2|*|x|=x^2|x|

Answer 12

x^2>=0*

Answer 13

yes, I know that;)

Answer 14

so I took x^2 out correctly?

Answer 15

yes
x^2/x^2=1

Answer 16

I mean without dealing with imaginary values.

Answer 17

because then I wouldn't, because
|-i| doesn't equal -i

Answer 18

but for real numbers, I took out the x^2 from top and bottom correctly... I suppose.

Answer 19

I think that since I can re-write your function as below:
f(x)= x^3 if x>=0
and
f(x)=-x^, x<0,
then the derivative is: