2006-10-05 20:10:38 · 9 answers · asked by pavan kumar NC 2 in Science & Mathematics ➔ Mathematics
I want step by step solution yar please....
2006-10-05 20:15:58 · update #1
0?
2006-10-05 20:11:47 · answer #1 · answered by Proverbs 2 · 0⤊ 1⤋
If you set f(0)=0 the function IS DIFFERENTIABLE in 0 as
(f(0+h)-f(0))/h=hsin(1/h) tends to 0 as h--->0
The derivative in 0 is f '(0)=0
Another thing is the continuity of the derivative in 0 (for JamesL). What you proove is that the derivative df/dx is not continuous in zero as f '(x) has no limit for x--->0
2006-10-05 20:43:37 · answer #2 · answered by 11:11 3 · 0⤊ 0⤋
Not differentiable.
Applying differentiation rules gives you 2x sin(1/x) - sin(1/x).
This function is undefined at x=0, and it does not have a limit as x->0 either, because of the second term.
2006-10-05 20:16:55 · answer #3 · answered by James L 5 · 1⤊ 0⤋
Equals 4
2006-10-05 20:11:52 · answer #4 · answered by Anonymous · 0⤊ 1⤋
It is not continuous at x = 0, hence not differentiable.
2006-10-05 22:29:16 · answer #5 · answered by Seshagiri 3 · 0⤊ 1⤋
the answer
2sin(1/x)*x-cos(1/x)
would be undefined at x=0
2006-10-05 22:07:41 · answer #6 · answered by Pam 5 · 0⤊ 0⤋
we use multiplication rule of differenciation
y= x^2sin(1/x)
dy/dx= x^2d/dx(sin(1/x)) + sin(1/x)d/dx(x^2)
dy/dx= x^2 cos(1/x)(-1/(x^2)) + 2xsin(1/x)
dy/dx at x=0
= 0
2006-10-05 21:41:07 · answer #7 · answered by Nick 3 · 0⤊ 1⤋
They are different
2006-10-05 20:11:52 · answer #8 · answered by Anonymous · 0⤊ 1⤋
hih
2006-10-06 06:00:13 · answer #9 · answered by janu 3 · 0⤊ 1⤋