Question

How to prove that at x=1 |x-1| is not differentiable?

Answer 1

is that y=

Answer 2

Yes! y=|x-1|

Answer 3

y=x-1
and y=-(x-1)
from the difintion of the absolute value

Answer 4

Is this the derivative?

f'(x)= { x-1, x>1
1-x, x<1

Answer 5

If so, how do I use the right and left hand derivatives to show that the derivative DNE?

Answer 6

the right derivative =1
the left derivative =-1

Answer 7

do you need to use limits here ?

Answer 8

Yes!

Answer 9

differentiable if
\(\large \lim \limits_{h \to 0} \dfrac{f(c+h)-f(c)}{h}\)
exists

Answer 10

so, take the left hand limit for this
our 'c' = 1
for left hand limit, you will take the function of x>1 or x< 1 ?

Answer 11

Is it x<1?

Answer 12

yes,
so our f(x) = 1-x
f(c+h) =... ?

Answer 13

f(c) = f(1) = 0 in both cases

Answer 14

Oh ok I see! Can I write Lim x--> 0 f'(x)?

Answer 15

Lim x--> 0 f'(x) for what ?

Answer 16

f'(x) itself is not defined...

Answer 17

I see..how would you do this without newton's quotient and using chain rule instead?

Answer 18

i am guessing,
f(x) = |x-1|
f'(x) = (|x-1|)' d/dx (x-1)
but since d/dx |x| does not exist , (|x-1|)' also DNE
so, f'(x) DNE

Answer 19

Isn't the derivative of |x-1|
x-1, x>1
1-x, x<1

Answer 20

I think?

Answer 21

the "definition "
of |x| is
x when x>0
-x when x<0
so what u said is the definition, not the derivative

Answer 22

Oops! I meant
1, x>1
-1, x<1?

Answer 23

thats correct

Answer 24

since they are not = at x=1
derivative DNE

Answer 25

but we didn't use chain rule here

Answer 26

Is chain rule Lim x--> c+ f'(x)?

Answer 27

what i know of chain rule is
\([f(g)]' = f'(g)\times g'\)

Answer 28

Sorry, bad terminology, I meant using derivatives, not newton's!

Answer 29

yeah, we used derivatie and proved that it didn't exist,
as left hand derivative (-1) not = to right hand derivative (+1), derivative at x=1 DNE

Answer 30

Ok I get it! Thank you!

Answer 31

welcome ^_^