Question

\(f(x) =\begin{cases}x^rsin(1/x) ~~if~~x>0\\0~~if~~x\leq 0\end{cases}\)
find values of r such that f(x) is everywhere differentiable.
Please, help

Answer 1

@wio @freckles @mathmate @M4thM1nd

Answer 2

Suppose I know how to argue some steps and I get
\(rx^{r-1}sin(1/x)+x^rlnx cos(1/x)=0\)

Answer 3

that is either x =0 or \(-\dfrac{r}{xlnx}= cot (1/x)\) then???

Answer 4

I think you've written \[\Large \frac{d}{dx} \frac{1}{x} \ne \ln x \\ \Large \frac{d}{dx} x^{-1} = -x^{-2}=\frac{-1}{x^2}\]

Answer 5

I don't get what you mean, how it relates to r?

Answer 6

When you did the chain rule after the product rule on taking the derivative here:

\(rx^{r-1} \sin(1/x)+x^r \ln x \cos(1/x)=0\)

I'm saying it should be

\(rx^{r-1} \sin(1/x)+x^r (-x^{-2}) \cos(1/x)=0\)

Answer 7

oh yeah, I see my mistake

Answer 8

thanks a lot, let me redo, I may get something else.

Answer 9

@Kainui I got it, hehehe... thanks a lot.

Answer 10

Cool =D