Question

one more, prob. Derivative of an abs value.... (remember that I am the asker:D )

Answer 1

\[\frac{d}{dx} \left(\begin{matrix} ~\left| \frac{x^2+x }{x-3}\right|~\end{matrix}\right)\]

Answer 2

I was typing, and then... dk

Answer 3

\[\dfrac{d}{d\clubsuit} (\left|\clubsuit\right|) = \dfrac{\clubsuit}{\left|\clubsuit\right|}\]

Answer 4

no I know how to do it, but I need to refresh

Answer 5

\[\frac{d}{dx} \left(\begin{matrix} ~\left| \frac{x^2+x }{x-3}\right|~\end{matrix}\right)=\frac{1}{2\left| \frac{x^2+x }{x-3}\right| }\] and multiply this times the derivative of the inside of the abs value.

Answer 6

the derivative of the inside is:
\[-(x^2+x)(x-3)^{-2}+(x-3)^{-1}(2x+1)\]

Answer 7

I will rearrange the derivative of the inner fraction, and re-write the entire thing

Answer 8

hold up

Answer 9

By chain rule you should get

\[\frac{d}{dx} \left(\begin{matrix} ~\left| \frac{x^2+x }{x-3}\right|~\end{matrix}\right)=\frac{\frac{x^2+x }{x-3}}{\left| \frac{x^2+x }{x-3}\right| }\cdot \frac{d}{dx} \left(\begin{matrix} ~\frac{x^2+x }{x-3}~\end{matrix}\right)\]


right ?

Answer 10

\[-(x^2+x)(x-3)^{-2}+(x-3)^{-1}(2x+1)\]\[-(x^2+x)(x-3)^{-2}+(x-3)^{-2}(2x+1)(x-3)\]\[(x-3)^{-2}[(2x+1)(x-3)-x^2-x]\]\[(x-3)^{-2}[2x^2-5x-3-x^2-x]\]\[(x-3)^{-2}[x^2-6x-3]\]\[So,~~\frac{d}{dx}~\frac{x^2+x}{x-3}=\frac{x^2-6x-3}{(x-3)^2}\]

Answer 11

why do you have
x^2+x
------
x-3

on top of the abs value?

Answer 12

because thats what the derivative of absolute value function says
incase if u have missed my first reply, here it is :
\[\dfrac{d}{d\clubsuit} (\left|\clubsuit\right|) = \dfrac{\clubsuit}{\left|\clubsuit\right|}\]

Answer 13

what I am trying to do is:\[|u|'=\frac{1}{2|u|}\times u'\]

Answer 14

isn't that so?

Answer 15

that doesnt look right. i think it should be like this : \[|u|'=\frac{u}{|u|}\times u'\]

Answer 16

your derivative for inside function is correct though

Answer 17

let see.
I am using

|u| = sqrt(u^2)
differentiating the entire sqrt argument:

the result: 1 / 2sqrt{u^2} or, 1 / ( 2|u| )

I see my mistake

Answer 18

I see.. you forgot to multiply by the derivative (u^2)' thats all :)

Answer 19

If I am defining |u| as sqrt(u^2),
then I would get to take the derivative of u^2:
so I get

2u (differentiating the entire u, by the power rule) and then, times the derivative of u.

Answer 20

so I would be getting:

|u| ' = sqrt{u^2} = 1/(2|u|) * (2u^(2-1)) * u'

Answer 21

I understand

Answer 22

so there the 2 cancel, and I get:

|u| ' = (u / |u| ) * u'

or, |u| ' = (u * u' ) |u|

Answer 23

yes, I understand your setup

Answer 24

looks good!

Answer 25

I can't copy my own code, I will refresh

Answer 26

\[\frac{d}{dx}~|\frac{x^2+x}{x-3}|=\frac{\frac{x^2+x}{x-3}}{|\frac{x^2+x}{x-3}|}\times \frac{x^2-6x-3}{(x-3)^2}\]

Answer 27

let me think if I can reduce this

Answer 28

\[\frac{d}{dx}~|\frac{x^2+x}{x-3}|=\frac{x^2+x}{|\frac{x^2+x}{x-3}|}\times \frac{x^2-6x-3}{(x-3)}\]step 1

Answer 29

technically you're done with finding the derivative
reducing is just algebra.. not calculus.. it shouldn't matter i guess

Answer 30

well, when usually I am required to reduce the simplest terms.

Answer 31

but I think I can comprehend this task.

Answer 32

hey wait

Answer 33

yes>?

Answer 34

careful with the fractions

Answer 35

I did it correctly (when I said step 1)

Answer 36

\[\begin{align}\frac{d}{dx}~|\frac{x^2+x}{x-3}|&=\frac{\frac{x^2+x}{x-3}}{|\frac{x^2+x}{x-3}|}\times \frac{x^2-6x-3}{(x-3)^2} \\~\\
&=\frac{x^2+x}{(x-3)|\frac{x^2+x}{x-3}|}\times \frac{x^2-6x-3}{(x-3)^2} \\~\\
\end{align}\]

Answer 37

I see ...

Answer 38

you are dividing by that (x-3) it is like in denominator

Answer 39

sorry

Answer 40

maybe think of it as multiplying top and bottom by (x-3)

Answer 41

\[\frac{d}{dx}~|\frac{x^2+x}{x-3}|=\frac{x^2+x}{|\frac{x^2+x}{x-3}|}\times \frac{x^2-6x-3}{(x-3)^3}\]

Answer 42

yes, I see what I did... stupid of me

Answer 43

-:(

Answer 44

\[\large \dfrac{~~\frac{a}{b}~~}{c} = \dfrac{a}{b\times c}\]

Answer 45

but basically, I can remember that:

\[\frac{d}{dx}~\left| f(x) \right|~=~\frac{f(x) \cdot f'(x)}{\left| f(x) \right|}\]

Answer 46

yes, ganesh, I undertstood tnx

Answer 47

\[\large \dfrac{~~\frac{a}{b}~~}{c} = \dfrac{~~\frac{a}{b}\times b~~}{c\times b }= \dfrac{a}{b\times c}\]

Answer 48

yes that will allow you to work with derivatives of ANY absolute value functions

Answer 49

yes....... you already showed me, lol

Answer 50

yes, tnx for formula verification

Answer 51

never intended to be a mathematician.... :)
tnx.

I think I am done with this problem.

Answer 52

putting up a tutorial ? just kidding...

Answer 53

Hey! I just wanted to say something for fun (and I know this sorta off topic and sorta on)
\[ \text{ Let } k \text{ be an integer } \\ \frac{d}{dx} |x^{2k}|= \frac{d}{dx} x^{2k}=2kx^{2k-1} \]
\[\frac{d}{dx} |x^{2k+1}|=\frac{x^{2k+1}}{|x^{2k+1}| } (2k+1)x^{2k}=\\ \frac{x^{2k}x}{|x^{2k}x|}(2k+1)x^{2k}=\frac{x^{2k}x}{x^{2k}|x|}(2k+1)x^{2k}=\frac{x}{|x|}(2k+1)x^{2k} \\ =\frac{|x|}{x}(2k+1)x^{2k}=|x|(2k+1)x^{2k-1}=(2k+1)x^{2k-1}|x|\]
just wanted to generalize the simplifcation @SithsAndGiggles was talking about

Answer 54

yes, tnx

Answer 55

\[\left|x^{2k+1}\right| = x^{2k} \left|x\right|\]

Answer 56

lol

Answer 57

stop making things easier

Answer 58

{o><o}
<{ }>
" "

Answer 59

"math to make things easier, not the opposite" or something along this...

Answer 60

I hold like that too.....

I feel like I am spending too much time on a single question.... I got my help, so will practice more.

Answer 61

\[[|x^{2k+1}|]'=[x^{2k}|x|]'=2kx^{2k-1}|x|+x^{2k} \frac{x}{|x|} =2kx^{2k-1}|x|+x^{2k} \frac{|x|}{x} \\ =|x|[2kx^{2k-1}+x^{2k-1}]=|x|(2k+1)x^{2k-1}\]

Answer 62

lol ifeel idku is more inclined towards square roots
\[\sqrt{(x^{2k+1})^2} = x^{2k}\left|x\right|\]

Answer 63

inclined is a very negative word

Answer 64

anyways, tnx again...
I will have to abandon this post.

Answer 65

basically you're not allowing the square root to evaluate to a negative number

Answer 66

I don't need any more stuff, I understand that:
\[\frac{d}{dx}~\left| f(x) \right|~=~\frac{f(x) \cdot f'(x)}{\left| f(x) \right|}\]
and actually, why is that...

so I am done with no other info.... tnx for assisting me this time

Answer 67

Fair enough! and I agree that formula is all you ever need if you're formula oriented :)