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Is this function differentiable at $\frac{\pi}{3}$?

$f: \mathbb R \to \mathbb R$, $f(x) \begin{cases} \tan x&\quad\text {if } -\frac{\pi}{2} < x < \frac{\pi}3\\ x^2&\quad\text{if } x \ge \frac{\pi}{3}\\ \end{cases}$

If it is differentiable at $\frac{\pi}{3}$, then show it.

I did this from the + side, and my result was $x_0 +\frac{\pi}{3}$, just need help on working $\tan x$, by the definition of derivative.

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    If you take the derivative of this function and approach it from the left, you get $4$. If you approach it from the right you get $\frac{2\cdot pi}{3}.$ – 2017-01-11

4 Answers 4

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The function is not differentiable at $\pi/3$, because it's not continuous at that point. Evaluating the first part at $\pi/3$ gives a result of $\sqrt 3$, while the second part gives $\pi^2/9$.

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Let's check continuity: letting $x_0 = \frac{\pi}{3}$, we have $$\lim_{x\to x_0^-}f(x) = \lim_{x\to x_0^-}\tan x = \sqrt 3 \neq \lim_{x\to x_0^+} f(x) = \lim_{x\to x_0^+} x^2 = \frac{\pi^2}{9}$$ Thus your $f$ is not continuous at $x_0$, which means it can't be differentiable there.

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Concisely, to check continuity, we evaluate $$ \lim_{x\rightarrow \pi/3}\tan(x)=\sqrt{3}\ne \pi^2/9=f(\pi/3) $$

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    Don't delete it based on my perception. I would suggest, however, that we both delete our comments above. – 2017-01-11
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As commented before $f$ is not continuos at $x=\pi/3$ so, it is not differentiable in that point. But if we want to find left and right derivative: $$f_{+}'(\pi/3)=\lim_{h\to 0^+}\frac{f(\pi/3+h)-f(\pi/3)}{h}=\lim_{h\to 0^+}\frac{(\pi/3+h)^2-(\pi/3)^2}{h}$$ $$=\lim_{h\to 0^+}\frac{(2\pi/3)h+h^2}{h}=\lim_{h\to 0^+}(2\pi/3+h)=2\pi/3.$$

$$f_{-}'(\pi/3)=\lim_{h\to 0^-}\frac{f(\pi/3+h)-f(\pi/3)}{h}=\lim_{h\to 0^-}\frac{\tan(\pi/3+h)-(\pi/3)^2}{h}.$$ But $\tan(\pi/3+h)-(\pi/3)^2\to \sqrt{3}-(\pi/3)^2\ne0$ as $h\to 0^-$, so $f_{-}'(\pi/3)$ does not exist.

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    In the second derivative, $f(\pi/3) = \tan(\pi/3),$ not $(\pi/3)^2$! See this line in your post: $$f_{-}'(\pi/3)=\lim_{h\to 0^-}\frac{f(\pi/3+h)-f(\pi/3)}{h}=\lim_{h\to 0^-}\frac{\tan(\pi/3+h)-(\color{red}{\pi/3)^2}}{h}.$$ – 2017-01-11
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    $\pi/3\ge \pi/3$, so $f(\pi/3)=(\pi/3)^2$ – 2017-01-11
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    I highlighted the problem. As h approaches from the left we need $$\frac{\tan((\pi/3) + h) -\color{blue}{ \tan(\pi/3)}} h$$ – 2017-01-11
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    Wrong. See my previous comment. – 2017-01-11
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    We are talking about the $f(x) = \tan x$, here. – 2017-01-11
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    In the first limit, approaching from the right, you are correct; it is the second limit, (in your post) approaching from the left that the relevant function is $f(x) = \tan x$. – 2017-01-11
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    So what? What is $f(\pi/3)$? Look again at the definition of $f.$ – 2017-01-11
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    @Alone. That square on that $(\pi/3)^2$ in the left side derivative needs to go. The formula is $y=tanx$, not $y=tan^2x$ – 2017-01-11
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    $f(\pi/3)=$? Please, answer. – 2017-01-11