Let f(x)=x2sin1xf(x)=x2sin1x for x≠0x≠0 and f(0)=0f(0)=0.
(a) Use the basic properties of the derivative, and the Chain Rule to show that ff is differentiable at each a≠0a≠0 and calculate f′(a)f′(a).
You may use without proof that sinsin is differentiable and that
sin′=cossin′=cos.
Not even sure what this is asking.
(b) Show that ff is differentiable at 00 and that f′(0)=0f′(0)=0.
f(x)−f(0)x−0→limx→0xsin(1/x)f(x)−f(0)x−0→limx→0xsin(1/x).
xsin(1/x)≤|x|xsin(1/x)≤|x| and limx→0|x|=0limx→0|x|=0.
Thus f(x)f(x) is differentiable at 00; moreover f′(0)=0f′(0)=0.
(c) Show that f′f′ is not continuous at 00.
f′(x)=x2cos(1/x)(−x−2)+2xsin(1/x)f′(x)=x2cos(1/x)(−x−2)+2xsin(1/x).
In pieces: limx→0cos(1/x)limx→0cos(1/x).
f′(0−)f′(0−) nor f′(0+)f′(0+) exists as x→0x→0 f′(x)f′(x) oscillates infinity between −1−1 and 11 with ever increase frequency as x→0x→0 for any p>0p>0 [−p,0][−p,0], [−p,p][−p,p] or [0,p][0,p] ff is not continuous.
Question: How to show more rigorously?
