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The condition of a derivative existing at a point is that slopes to the curve drawn from the left hand side of the point, and the slopes drawn from right hand side of the point must approach the same value. The condition for continuity at a point is that the value of a function must must approach the same value from both sides as we approach the point from both sides.

The value of the derivative function is the slope itself, and the derivative function exists at a point only when the slope approaches the same value from both sides (left hand derivative = right hand derivative condition). That means it's at the point continuous too right? Is there anything wrong with this reasoning?

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  • $\begingroup$ No. See math.stackexchange.com/a/423279/551592 However, although derivative doesn't need to be continuous, it still has IVP: en.wikipedia.org/wiki/Darboux%27s_theorem_(analysis) $\endgroup$ Nov 2, 2019 at 23:26
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    $\begingroup$ A correct statement is that if the derivative $f'(x)$ exists at a point $c$, then the function we differentiated $f(x)$ is continuous at that point. Did you mean that? The derivative is a separate function which may or may not be continuous at $c$. $\endgroup$
    – bjorn93
    Nov 2, 2019 at 23:30
  • $\begingroup$ @bjorn To me it seems like the condition LHD= RHD for the existence of derivative is very similar to the condition for the continuity of the derivative. I don't know what I'm missing here. $\endgroup$
    – Ryder Rude
    Nov 3, 2019 at 3:05
  • $\begingroup$ @RyderRude The issue is that when you say the derivative is continuous, you're saying something about the derivative as a separate function. The existence of the derivative at a point doesn't give us enough information. $\endgroup$
    – bjorn93
    Nov 3, 2019 at 3:10
  • $\begingroup$ @bjorn93 This is how it looks in my head 'If a function is differentiable at a point, then the slopes drawn to the left of the point and the slopes drawn to the right of the point approach the same value as we approach the point (This is the LHD=RHD condition). We define this same value as the derivative at that point. If we now plot the graph of the slope values around that point (graphing the derivative),then we should expect the values to the right of the point and those to the left of the point to again approach the same value (that's the reason we said the derivative existed at the point) $\endgroup$
    – Ryder Rude
    Nov 3, 2019 at 3:18

3 Answers 3

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Consider the function:

g(x) := \begin{array}{rl} x^2 \sin(1/x), & x \ne 0 \\ 0, & x = 0 \\\end{array}.

Then, we have:

g'(x) = \begin{array}{rl} 2x \sin(1/x) - \cos(1/x) & x \ne 0 \\ 0, & x = 0 \\\end{array}

Observe that $g'$ is not continuous at $0$ but $g'(0)=0$.

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"The value of the derivative function is the slope itself, and the derivative function exists at a point only when the slope approaches the same value from both sides (left hand derivative = right hand derivative condition). That means it's at the point continuous too right? Is there anything wrong with this reasoning?"

Not quite, you are indeed describing in words the math $$ \lim_{x\to x_0^+}f'(x)=\lim_{x\to x_0^-}f'(x) $$ But that's not what you check when you compute a derivative: There you are taking limits of secant lines from both sides, which must agree. This is the math $$ \lim_{x\to x_0^+}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0^-}\frac{f(x)-f(x_0)}{x-x_0} $$

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To add to the examples in the other answers: the function given by $f(x)=x^2\mathbb 1_{\mathbb Q}(x)$, (that is, which vanishes on the irrationals and is the squaring function on the rationals) is continuous at $0$ and differentiable at $0$, but not continuous nor differentiable anywhere else.

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