Proving Differentiability of W(x) in the Context of U(x) = 2 + x^2 + x*W(x)

[omni]

1)be W(x) Continuous function for any X.
U(x)=2+x^2+x*W(x)

Prove that U(x) are Differentiable in X=0

well i know that i need to Prove that W(x) is also Differentiable but how i Prove it?

thank you.

 

[HallsofIvy]

You don't! You have no control over "W". You are told that W is continuous and that is all you can assume about W. If you believe that W must also be differentiable for the conclusion to be true then you believe that the statement you are asked to prove is false!

For example, suppose W(x)= |x| which is continuous for all x but not differentiable at x= 0. Then U(x)= 2+ x^2+ x|x|. If x> 0, that is the same as U(x)= 2+ x^2+ x^2= 2+ 2x^2. That has derivative U&#039;= 4x which goes to 0 as x goes to 0. If x< 0, that is the same as U(x)= 2+ x^2- x^2= 2 which has derivative 0 for all x> 0. Again that goes to 0 as x goes to 0.

That requires knowing that a function, f, is differentiable at x= a if and only if \lim_{x\to a^-}f&#039;(x)= \lim_{x\to a^+}f&#039;(x) which is a result of the fact that, even though a derivative is not necessarily continuous it must still satisfy the "intermediate value property".

More direct would be to calculate the derivative at 0 directly from the definition:
\lim_{h\to 0}\frac{U(h)- U(0)}{h}= \lim_{h\to 0}\frac{(2+ h^2+ h|h|)- 2}{h}

\lim_{h\to 0^+}\frac{h^2+ h|h|}{h}= \lim_{h\to 0^+}\frac{2h^2}{h}= \lim_{h\to 0^+}2h= 0
\lim_{h\to 0^-}\frac{h^2+ h|h|}{h}= \lim_{h\to 0^-}\frac{0}{h}= \lim_{h\to 0^+}0= 0
Since those two limits exist and are equal,
\lim_{h\to 0}\frac{h^2+ h|h|}{h}
itself exists and the function is differentiable.

Do that with U(x)= 2+ x^2+ xW(x) using only the fact that W is continuous.

 

[tiny-tim]

hi omni!

(have a delta: δ and an epsilon: ε and try using the X² tag just above the Reply box )

(i assume you mean that it is differentiable at x = 0 ?)

no, just use the δ,ε definition of continuous at x = 0 to show that xW is differentiable at x = 0 …

what do you get? ​

 

[╔(σ_σ)╝]

Mm

hi omni!

(have a delta: δ and an epsilon: ε and try using the X² tag just above the Reply box )

(i assume you mean that it is differentiable at x = 0 ?)

no, just use the δ,ε definition of continuous at x = 0 to show that xW is differentiable at x = 0 …

what do you get? ​

xW(x) need not be differentiable at zero. Hallsofivy already gave a good counter example.

 

[tiny-tim]

Mm

xW(x) need not be differentiable at zero. Hallsofivy already gave a good counter example.

no, he gave a good example of it being true

 

[╔(σ_σ)╝]

Oops. I guess I should read more carefully next time. Sorry about that; please carry on!

@OP

The fact that as x->0 the function g(x) =x goes to zero will give you a feel for why this is true. Also since in every delta neighbourhood of 0, W(x) is bounded it is not impossible that xW(x) goes to zero as x goes to zero.

 

[omni]

hi thanks to all of u about the answers.
i did like this (i diden't Write the lim and h-->0 but i know is need to be there)
f' = [U(x+h)-U(x)]/h = {[2+(x+h)² + (x+h)*Z(x+h)] - [2+x²+x*Z(x)]}/h =

[2+x²+2hx+h²+x*Z(x+h)+h*Z(x+h) - 2-x²-x*Z(x)]/h = [2hx+h²+x*Z(x+h)+h*Z(x+h)-x*Z(x)] / h =

= 2x+h+[(x+h)*Z(x+h) - x*Z(x)]/h
how i keep on from here?

thanks