y = |x|/x

To the right of the y-axis I think would be the line y = 1. To the left of the y-axis would be the line y = -1. Undefined at x = 0.

Not sure though...

Hope this helps,

~~Simba

Reply 4

Dekota

Undefined at x = 0,
for -ve values of x, y = -1,
for +ve values of x, y = 1

Is discontinous.

Reply 5

Toffee_Girl

Simba

y = |x|/x

To the right of the y-axis I think would be the line y = 1. To the left of the y-axis would be the line y = -1. Undefined at x = 0.

Not sure though...

Hope this helps,

~~Simba

Thats great, Thanks!

Reply 6

topcrabcakes

V.P. Keys

dy/dx = |x|/x

It should be:

dy/dx=x/|x|

Reply 7

Dekota

wacabac

It should be:

dy/dx=x/|x|

^o) Okay ... although I don't see the difference between the two ...

Reply 8

topcrabcakes

None really, I'm being an idiot :wink:

Reply 9

Dekota

I learnt the result from Neopoliton.

:cool:

Reply 10

Spunky

Never came across this one when I was doing A-Level, does anyone know the actual steps to get to the term for dy/dx :smile:

Reply 11

Dekota

Casey

Never came across this one when I was doing A-Level, does anyone know the actual steps to get to the term for dy/dx :smile:

No definitive way of differentiating modulus functions, it's just by inspection.

Reply 12

Spunky

V.P. Keys

No definitive way of differentiating modulus functions, it's just by inspection.

Cool

, at least I'm not missing out on some fundamental first principles knowledge :biggrin:

Reply 13

Neapolitan

Casey

Cool

, at least I'm not missing out on some fundamental first principles knowledge :biggrin:

It's just the same as you'd differentiate x and -x separately and x/|x| is a shorthand way of writing your answer

Reply 14

topcrabcakes

Is it incorrect to say:

y=|x|

=> y²=x²
=> 2y(dy/dx)=2x
=> dy/dx=x/y

dy/dx=x/|x| ?

Reply 15

Neapolitan

Yes that's a neat way to do it as well and valid when x≠0.

Reply 16

Spunky

wacabac

Is it incorrect to say:

y=|x|

=> y²=x²
=> 2y(dy/dx)=2x
=> dy/dx=x/y

dy/dx=x/|x| ?

Neat

and yeah Neapolitan, I think I'd be able to do it if it came up in a question by considering both sides, but its just something that I hadn't come across though that probably has more to do with missing all the maths lectures this term :p:

Reply 17

mathsexams

Casey

Never came across this one when I was doing A-Level, does anyone know the actual steps to get to the term for dy/dx :smile:

Write it as |x| ≡ √(x²) and use the chain rule.

You can use this to get the second derivative too :smile:

Reply 18

Spunky

mathsexams

Write it as |x| ≡ √(x²) and use the chain rule.

You can use this to get the second derivative too :smile:

I'll try it that way too :smile:

, actually if the working isn't too convoluted (which from first impressions it doesn't look like it will be) then I'd probably prefer that method as it doesn't involve having to remember to use a substitution.

Reply 19

Trzylraj

=x/|x|

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