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A question defines $$f(x)=\begin{cases}g(x)\cos\frac{1}{x}&\text{if } x\neq0\\0&\text{if } x=0\end{cases}.$$ Here, $g$ is an even function and differentiable at $0$, and $g(0)=0$. You're required to find $f'(0)$.

On differentiating $f(x)$ to find it, you get $$g'(x)\cos\frac{1}{x}+\frac{g(x)\sin\frac{1}{x}}{x^2}$$. The first term is obviously $0$ at $x=0$, but the second term is in a $\frac{0}{0}$ indeterminate form, as both $g$ and $x^2$ approach 0 at x=0. But all the solutions I've seen simplify that term to zero, too.

Isn't that incorrect? I tried using L'Hôpital's Rule on $\frac{g(x)}{x^2}$, but since we don't know what g'(x) is, it doesn't help much. So how does that term end up as 0?

harry
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    Note that $f'(0)$ isn't necessarily equal to $\lim_{x \to 0} f'(x)$, so your method is incorrect, you have to apply the definition directly to find $f'(0)$. – Hans Lundmark Nov 20 '20 at 06:48

2 Answers2

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Since $g$ is even and differentiable at $0$ we must have $g'(0)=0$.

All you have to do is apply the definition of derivative. $f'(0)=\lim_{x \to 0} \frac {f(x)-f(0)} x=\lim_{x \to 0} \frac {g(x)} x \cos (\frac 1x)=0$ because $ \frac {g(x)} x \to 0$ and $\cos (\frac 1 x)$ is bounded.

Kavi Rama Murthy
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  • I got it; for further reference, I'm leaving an explanation for your first point here. Also, you can test the differentiability at a point using either the first principle or by finding the differential of the function as usual, right? – harry Nov 20 '20 at 06:55
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    @HarryHolmes $g$ is only given to be differentiable at $0$. $g'(x)$ may not exist for any $x \neq 0$. So your approach fails. – Kavi Rama Murthy Nov 20 '20 at 07:21
  • That's right, but in cases where we do know the function, wouldn't it be apparent when I use direct differentiation? For example, while evaluating f(x) =|x|, it's clear when you attempt to differentiate it that it isn't differentiable at 0. In this case, we don't know what the function is, so you need to use first principle. But isn't the direct method okay in normal circumstances? – harry Nov 20 '20 at 09:50
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You have that $f(0)=0, f(t)=f(-t)$, which implies graphically that $x=0$ is a turning point of $f$, where its derivative is zero.

You can check this is the case, as the derivative calculated from both sides is equal. In the positive case:

$$f'(0)=\lim_{t\to0}\frac{f(t)-f(0)}{t}=\lim_{t\to 0}\frac{f(t)}{t}$$ and in the negative case: $$f'(0)=\lim_{t\to 0} \frac{f(-t)-f(0)}{-t}=\lim_{t\to 0}-\frac{f(t)}{t}\overbrace{=}^{\text{ AOL}}-\lim_{t\to 0}\frac{f(t)}{t}$$

Since $f'(0)=-f'(0)$, we have that $f'(0)=0$.

Rhys Hughes
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