Question

Another derivative (see inside)

Answer 1

Find \(f^\prime(0)\) if
\[
f(x) = \left\{
\begin{array}{lr}
g(x)\sin\frac{1}{x}, & x \not=0\\
0, & x =0
\end{array}
\right.
\]
and
\[
g(0)=g^\prime(0)=0
\]

Answer 2

I found the derivative to be
\[
f^\prime(x)=g^\prime(x)\sin{\frac{1}{x}}-\frac{g(x)\cos\frac{1}{x}}{x^2}
\]
but if you plug in zero, you end up dividing by zero all over the place, which I think is kind of the point, but I don't know what I'm supposed to do with this.

Answer 3

well the 1st term is 0 since g'(0) = 0
for 2nd term , i think you need to change the limit
let n = 1/x
\[f'(0) = \lim_{n \rightarrow \infty}-\frac{g(1/n) \cos n}{1/n^{2}}\]
this may work , not sure though

Answer 4

nope all i get is infinity...doesn't seem to work

Answer 5

And of course this one also does not have an answer in the back of the book haha. I don't even know where to start with this. I tried looking at it from the limit definition of the derivative and that didn't help at all. I don't have much experience with differentiating piecewise functions.

Answer 6

my guess is limit does not exist and f'(0) is undefined

http://www.wolframalpha.com/input/?i=lim+sin%281%2Fx%29-+%28cos%281%2Fx%29%2Fx%29+as+x-%3E0

Answer 7

That's a really unsatisfactory answer, especially if I can't give a good reason for it being the answer.

Answer 8

Also I don't get why he would ask the question if that was it

Answer 9

Also we have to make use in some way of the fact that f(0)=0

Answer 10

my guess is that the derivative of f′(0) is non-existed because it is a zero

Answer 11

that is true...i just don't know the answer
if you assume g(x) is some polynomial then from the graph, it shows clear evidence of why f(x) is not differentiable at 0

http://www.wolframalpha.com/input/?i=x*sin%281%2Fx%29+from+-.001+to+.001

Answer 12

Alright, I give up on this one. Which is terrible because this is problem 11, and there are 35 problems in this chapter. Oh well, on to the next chapter.