When f is the function defined by?

Question

f(x)= { x sin 2/x , x not equal to 0
{ 0 , x=0
determine if f '(0) exists and if it does, find its value

how would i do this? can you show me the easiest way to understand this? thanks

Answer 1

it's asking you to see if it has a derivative

in this case you'd use the multiplication rule w/ derivatives...and do

(X * the derivative of sin (2/x)) + (sin (2.x) * 1)

since the derivative of x = 1...you'd use a 1 in the second set of parenthesis

then you'd just simplify that out...i think the derivative of sin is cosine....

good luck

Answer 2

Since the function does not exist at x = 0, it does not have a derivative at x =0.

Answer 3

First you got to differentiate it

f(x)=xsin2x
and find f'(0)

okay how to differentiate it, first you have to use the v'u+u'v rule

so its
f'(x)= x . 2.cos2. 0 + 1.sin2x

f'(0) = 0. 2cos2.0 + 1.sin 2.0

so f'(0) = 0

Which actually makes a lot of sense when you think about it

When you differentiate sinx so you get cosx. When the cos graph equals zero it has a gradient of 0 too