Differentiability of a functionWatch

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Thread starter 14 years ago
#1
Just another question that's been annoying me:

Prove that the function

f(x) := {[0 if x=0],[(x^2)*sin(1/x) otherwise]}

is differentiable for all x in R.

thanks
0
14 years ago
#2
using product rule where u = x² and v = sin(1/x)
du = 2x and dv = cos 1/x * -1/x²
-(cos1/x)*x²/x² + 2xsin(1/x)

=2x.sin(1/x) - cos(1/x)

Thats not a proof but it is the answer when you differentiate it
0
14 years ago
#3
(Original post by manps)
using product rule where u = x² and v = sin(1/x)
du = 2x and dv = cos 1/x * -1/x²
-(cos1/x)*x²/x² + 2xsin(1/x)

=2x.sin(1/x) - cos(1/x)

Thats not a proof but it is the answer when you differentiate it
That's all well and good for x =/= 0, but you need to can't apply the chain/product rules at x=0.

At x=0 you need to note;

(f(x)-f(0))/(x-0) = xsin(1/x) ->0 as x->0.
0
Thread starter 14 years ago
#4
(Original post by RichE)
That's all well and good for x =/= 0, but you need to can't apply the chain/product rules at x=0.

At x=0 you need to note;

(f(x)-f(0))/(x-0) = xsin(1/x) ->0 as x->0.
thanks, at least I've done that bit right!

but any ideas for a PROOF of differentiability when x =/= 0?
0
14 years ago
#5
(Original post by flyinghorse)
Prove that the function

f(x) := {[0 if x=0],[(x^2)*sin(1/x) otherwise]}
is differentiable for all x in R.
Let y = f(x) ---> y = x^2.sin(1/x)
Let u = x^2 ---> du/dx = 2x
Let v = sin(1/x) = sin(x^-1) and Let w = x^-1 ---> dw/dx = -1/x^2. Hence v = sinw ---> dv/dw = cosw.
Hence: dv/dx = dv/dw * dw/dx = cosw * (-1/x^2) = -[cos(1/x)]/x^2
Hence: f ' (x) = dy/dx = u(dv/dx) + v(du/dx) = 2xsin(1/x) - cos(1/x)

When x = 0: f(x) = 0 ---> f ' (x) = 0 ---> f ' (x) is defined for x = 0 ---> Function is differentiable when x = 0.
When x =/= 0: f(x) = (x^2)*sin(1/x) ---> f ' (x) = 2xsin(1/x) - cos(1/x) ---> f ' (x) is defined for x =/= 0 ---> Function is differentiable when x =/= 0.

As the function f(x) is differentiable when x = 0 and x =/= 0 ---> Hence the function f(x) is differentiable for all x in R.

QED.

Nima
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