Show that the function `f(x)={x^msin(1/x), 0 ,\ \ \ x!=0 ,\ \ \ \ \ \ \x=0` is differentiable at `x=0` , if `m >1`

Explanation

Given function
`f(x)={{:(x^msin(1/x), xne0),(0,x=0):}`
`LHD` at `x=0`
`lim_(x->0^-)(f(x)-f(0))/(x-0)`
`=lim_(h->0)(f(0-h)-f(0))/(0-h-0)`
`=>lim_(h->0)((-h)^msin(-1/h))/-h`
`=>lim_(h->0)((-h)^(m-1)sin(-1/h))`
`=0`
`RHD` at `x=0`
`lim_(x->0^+)(f(x)-f(0))/(x-0)`
`=lim_(h->0)(f(0+h)-f(0))/(0+h-0)`
`=>lim_(h->0)((h)^msin(1/h))/h`
`=>lim_(h->0)((h)^(m-1)sin(1/h))`
`=0`
clearly,`LHD=RHD`
hence, f(x) is differentiable at `x=0`
, if `m >1`