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Say the value of $dy/dx$ for a function $y=f(x)$ at $x=a$ was turning out to be some finite value $A$, is it still possible that the function is non-differentiable at $x=a$.

If I'm getting a finite value of derivative for a function at a point, can I be sure about that the function is differentiable at that point? or is it like getting a finite value at a point doesn't guarantee that the function is differentiable at that point.

Falcon
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Harshit Rajput
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  • Your question sounds like “If $f'(a)$ exists, can I be sure that $f$ is differentiable at $a$?” Is that what you mean? – José Carlos Santos Jan 15 '22 at 11:59
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    Maybe do you have a precise example in mind ? The question doesn't really make sens to me. – Falcon Jan 15 '22 at 12:01
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    @JoséCarlosSantos yes, except for exists I mean "is some finite value" – Harshit Rajput Jan 15 '22 at 12:04
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    Since, by definition, $f'(a)$ is the limit $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$ when this limit exists and it is a real number, your question makes no sense to. – José Carlos Santos Jan 15 '22 at 12:06
  • When you say "you are getting", you are getting a value how? I think you are confusing what it means to be not differentiable - for example, $|x|$ is not differentiable at $0$ even though we "get" two values $1,-1$ - it's not differentiable precisely because we are getting multiple values which don't agree with each other – FShrike Jan 15 '22 at 12:07
  • @FShrike My question stems from a function which was similar to |x|, i.e. it had a kink. Taking the example of |x|, I thought that we can take a value of x just after x=0 and just before x=0, and then take a difference of these quantities and then divide by the differential distance of these points, to get a finite value. So that is what I meant, that even though |x| is not differentiable at x=0 I can get a value of dy/dx at x=0 – Harshit Rajput Jan 15 '22 at 12:17
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    So, no. See this. – David Mitra Jan 15 '22 at 12:19
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    You are getting a value, not of $\mathrm{d}y/\mathrm{d}x$, but rather of some $\Delta y/\Delta x$ approximation. Furthermore, I think from your description that you are in fact taking a "symmetric derivative" which is not the same as a derivative - the derivative centres itself at $x_0$ and considers perturbations $x_0+h$; the symmetric derivative considers $x_0-h,x_0+h$ and this is not the same thing - e.g., $|x|$ is symmetrically differentiable everywhere but not differentiable everywhere – FShrike Jan 15 '22 at 12:19
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    @HarshitRajput " that even though |x| is not differentiable at x=0 I can get a value of dy/dx at x=0 " No this is not true,you cannot get a value of dy/dx at x=0 as Right hand derivative will be 1 where as Left Hand Derivative to be -1. –  Jan 15 '22 at 12:20
  • Understood, that value that I'm talking about is just a $\frac{\Delta y}{\Delta x}$ not dy/dx. Thank You All. – Harshit Rajput Jan 15 '22 at 12:33

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