differentiable, f' not continuous

**chloe561** · March 19th 2009, 02:46 PM

Let $f(x)=x^2 \sin(\frac{1}{x})$ for $x \not = 0$ and $f(0)=0.$

a) Prove that $f(x)$ is differentiable on $\mathbb{R}$ .
b) Prove that $f'(x)$ is not continuous on $\mathbb{R}$ .

I need help on BOTH parts. I've tried this problem, but I think I am not being very explicit in my proofs. I am also very sure my proof in b) is not nice. Is there another way to do b) [possibly without using the fundamental theorem of calculus]?

Attempt:
a) Clearly $f(x)$ is differentiable at $x \not = 0$ . [How do I show this?] In fact, $f'(x)=2x \sin(x) - \cos(\frac{1}{x})$ [valid for $x \not =0$ ]. At $x=0$ , $f'(0)=\lim_{x \rightarrow 0} \frac{f(h)-f(0)}{h}=\lim_{h \rightarrow 0} h \sin(\frac{1}{h})=0$ . Hence, $f(x)$ is differentiable on $\mathbb{R}$ .

b) $f'(x)=2x \sin(x) - \cos(\frac{1}{x})$ . Let $p(x)=2x \sin(x)$ . Then $p(0)=0$ because $p(0)=2 \cdot \lim_{x \rightarrow 0} x \sin(\frac{1}{x})=0$ . So $p(0)=0$ . So by FTC $\exists P(x)$ s.t. $P'(x)=p(x)$ . So $f'(x)=P'(x)-cos(\frac{1}{x})$ . But then $\cos(\frac{1}{x})$ is not a continuous function. But by FTC, $P(x)$ is differentiable. Then $P(x)$ , $f(x)$ are differentiable functions whose derivatives are not continuous.

Thanks.

**Krizalid** · March 19th 2009, 04:11 PM

$f'(x)\to\infty$ as $x\to0,$ thus $f'(x)$ is not continuous at $x=0.$ $f$ is of course, differentiable but not of class 1.

**Laurent** · March 21st 2009, 06:08 AM

Originally Posted by chloe561

Let $f(x)=x^2 \sin(\frac{1}{x})$ for $x \not = 0$ and $f(0)=0.$

a) Prove that $f(x)$ is differentiable on $\mathbb{R}$ .
b) Prove that $f'(x)$ is not continuous on $\mathbb{R}$ .

I need help on BOTH parts. I've tried this problem, but I think I am not being very explicit in my proofs. I am also very sure my proof in b) is not nice. Is there another way to do b) [possibly without using the fundamental theorem of calculus]?

For part a), to justify that $f$ is differentiable on $\mathbb{R}\setminus\{0\}$ , you can say that it is obtained by product and composition of the following differentiable functions: $x\mapsto x^2$ on $\mathbb{R}$ , $\sin$ on $\mathbb{R}$ and $x\mapsto\frac{1}{x}$ on $\mathbb{R}^*$ . The computation at 0 is fine (well, it would be better not to write $f'(0)$ when you don't know that it exists yet)

For part b), you don't need to refer to FTC : just say that $x\sin\frac{1}{x}$ tends to 0, while $\cos\frac{1}{x}$ has no limit as $x\to0^+$ , so that $f'(x)$ has no limit as $x\to0^+$ . Therefore, it can't be continuous (whatever $f'(0)$ is).

Originally Posted by Krizalid

$f'(x)\to\infty$ as $x\to0,$ thus $f'(x)$ is not continuous at $x=0.$ $f$ is of course, differentiable but not of class 1.

No, $f'(x)$ is bounded near 0 (and it has no limit when $x\to0^+$ ).

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