Question

Given the function, find a formula of f ' (x).

Answer 1

\[f(x) = x \left| x \right|\]

Answer 2

sare you trying to find the equation?

Answer 3

yes

Answer 4

Well, I think it would be Y=x|x| there isn't much of a way to reduce it any further.

Answer 5

the answer is \[2\left| x \right|\], so somehow it can be done. But, I don't know how to get it in that form. lol

Answer 6

Oh I see it now!

Answer 7

have you tried using product rule...
also do you know the derivative of |x|?

Answer 8

if you don't know the derivative of |x|
I can help you with that

Answer 9

Sorry if this is anyway confusing, I'm on mobile and it doesn't let me type out the equation.

1|x|+x (x/x)(x)

Which =2|x|

Answer 10

Okay, tell me please :)

Answer 11

\[\text{ Let } y=|x| \\ \text{ \square both sides } \\ y^2=|x|^2 \\ y^2=x^2 \\ \text{ differentiate both sides } \\ 2 y y'=2x \\ \text{ solve for } y' \\ y'=\frac{2x}{2y} \\ y'=\frac{x}{y} \\ \text{ recall } y=|x| \\ \text{ so } y'=\frac{x}{|x|}\]

Answer 12

you did beat me though @Astrophysics :p

Answer 13

To get rid of the absolute sign, you would need to square both sides. Yes?

Answer 14

Haha, I like your explanation though :P

Answer 15

yes @Zenmo

Answer 16

How would I get it to be in the form of \[2\left| x \right|\] ?

Answer 17

can you show me what you have after product rule

Answer 18

and there is a cute little trick we can do

after product rule:
combining fractions
|x|^2=x^2
and finally using that backwards x^2=|x^2|
then using that |x^2|/|x|=|x^2/x|=|x|

:) try using these hints after the product rule

Answer 19

\[y ' = x \times \frac{ x }{ \left| x \right| } + (1) \times \left| x \right|\]

Answer 20

what i mean by combining the fractions is writing the two terms you have there as one term

Answer 21

if you need another hint
you can do this by multiplying that second term you have there by |x|/|x|

Answer 22

\[\frac{ x^2 }{ \left| x \right| } + \frac{ \left| x \right|\left| x \right| }{ \left| x \right| }\]
\[\frac{ x^2+\left| x \right|\left| x \right| }{ \left| x \right| }\]

Answer 23

right and note that |x|*|x|=|x|^2=?

Answer 24

| x^2 |

Answer 25

or just x^2

Answer 26

why not -x^2?

Answer 27

x^2 is never negative

Answer 28

ah, I see.

Answer 29

absolute value doesn't change the result

Answer 30

\[\frac{x^2+x^2}{|x|}\]
I hope it is obvious why we wanted to write |x|^2 as x^2

Answer 31

if not obvious we wanted to do this because it would make it more clear we have like terms

Answer 32

in the numerator

Answer 33

i see

Answer 34

\[\frac{ 2x^2 }{ \left| x \right| }\]

Answer 35

right...
\[\frac{2x^2}{|x|}=\frac{2|x^2|}{|x|}=2|\frac{x^2}{x}|=...\]

Answer 36

Yea, I see it now. Thanks.
I don't have much knowledge of absolute signs manipulation.
:)

Answer 37

i found a faster way to do this problem:
\[y=x|x| \\ y^2=(x|x|)^2 \\ y^2=(x)^2(|x|)^2 \\ y^2=x^2x^2 \\ y^2=x^4 \\ \text{ differentiate both sides } \\ 2y y'= 4x^3 \\ y'=\frac{4x^3}{2y} \\ y'=\frac{2x^3}{y} \\ y'=\frac{2x^3}{x|x|} \\ y'=\frac{2x^2}{|x|}\]
and then the rest of the steps are the same

Answer 38

How would I create a piece-wise function of | x^2| ?

Answer 39

|x^2|=|x|^2=x^2
that is just a parabola
you could write a piece function though