Question

FINDING THE DERIVATIVE

F(x)=|1/x| that's an absolute value. I have a good knowledge of how to find derivatives but when I saw the absolute value I blanked out. I'd appreciate any help, thank you

Answer 1

In general, the derivative of an absolute value function
\(y=|f(x)|\) is given as \(y'=\displaystyle \frac{f(x)}{|f(x)|}\times f'(x)\).

Answer 2

So it would be 1 • 1/x ?

Answer 3

\(\displaystyle \frac{1/x}{|1/x|}\times \left(\frac{d}{dx} (1/x)\right)\)

Answer 4

if you go according to that.

Answer 5

You could simplify it in several ways, but what you did is not correct.

Answer 6

I understand but what happens with the absolute value in the denominator?

Answer 7

Thank you, SolomonZelman
It is really helpful for me!

Answer 8

Could you simplify it ?

Answer 9

I know it's 1/x / |1/x • 1/x^2

Answer 10

actually it is -1/x^2.

Answer 11

I think it will be
\[y'=\frac{ -1 }{ x^3|\frac{ 1 }{ x }| }\]

Answer 12

Yeah that's what I meant

Answer 13

\(\displaystyle \frac{1/x}{|1/x|}\times \left(\frac{d}{dx} (1/x)\right)=\frac{1/x}{|1/x|}\times \left(\frac{-1}{x^2}\right)=\)

\(\displaystyle (1/x)\times (|x|/1)\times\frac{-1}{x^2} =-\frac{|x|}{x^3}=-\frac{|x|}{|x^2|x}=\frac{-1}{|x|x}\).

Answer 14

Great!

Answer 15

I will give you the derivation of it.
------------------------------
PROOF (not quite rigorous, but....)

Let \(f\) be a function of \(x\), and let \(y=|f|\).
Assume \(f\) is differentiable, and assume \(x\in \mathbb{R}\).
(You'll see why the 2nd assumption is there.)

Definition of Absolute Value (for real numbers): \(z=\sqrt{z^2}\).
Rewriting the function this way gives \(y=\sqrt{f^2}\). (Don't forget, \(f\) is a function of \(x\).)

Applying the rules of differentiation:
\(\displaystyle y'=\frac{1}{2\sqrt{f^2}}\times \left[\frac{d}{dx}(f^2)\right]=\frac{1}{2\sqrt{f^2}}\times \left[2f\times f'\right]=\frac{f\times f'}{\sqrt{f^2}}\).

applying the definition of the absolute value, back,
\(\displaystyle y'=\frac{f\times f'}{|f|}\).

Answer 16

This is why / Thus .... \(\displaystyle\frac{d}{dx}|f(x)|= \frac{f(x)}{|f(x)|}\times f'(x)\).

Answer 17

One question though. Why did you make the x^3 into |x^2|x

Answer 18

And how

Answer 19

nice ques!

Answer 20

\(x^2=|x^2|\) for any real x. (Right?) Hence, \(x^3=x^2\times x=|x^2|x\).

Answer 21

or, do you mean by "why" - "to accomplish what did I do this" ?

Answer 22

I see. But why not make it into |x|x^2 ?

Answer 23

Nono i know why you did it

Answer 24

if \(x\) is negative, then \(|x|x^2\) is positive and \(x^3\) is negative.
They are (equivalent in magnitude, but) not equivalent in sign.

Answer 25

questions?

Answer 26

No i got it now, I was trying to work to out thanks!

Answer 27

Helped a lot

Answer 28

Tnx for the feed:)