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Somehow I'm getting the conclusion that the only curve which can satisfy this condition is a straight line. My reasoning is:

The right hand derivative of a function $f(x)$ at point $x=a$ is

$$\lim_{h\rightarrow 0^+}\frac{f(a+h)-f(a)}{h}=f'(a)$$

And, the left hand derivative of the function at a very close point $a+h$ ($h\rightarrow 0$) is:

$$\lim_{h\rightarrow 0^+}\frac{f(a+h-h)-f(a+h)}{-h}=\lim_{h\rightarrow 0^+}\frac{f(a+h)-f(a)}{h}=f'(a)$$

This means the right hand derivative of a function at a point $a$ equals the left hand derivative at point $a+h$ ($h\rightarrow 0$). Since the function is everywhere differentiable, so LHD at $a+h$ equals RHD at $a+h$. So, RHD at $a+h$ is also equal to $f'(a)$. Now, by above reasoning, RHD at $a+h$ equals LHD at $a+2h$. So, LHD at $a+2h$ is also $f'(a)$. By continuing this reasoning, the derivative of the function at all the points is $f'(a)$ which makes the curve a straight line.

Ryder Rude
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  • Essentially, your conception and definition of the left hand derivative is incorrect. See the page https://math.stackexchange.com/questions/1158510/left-hand-derivative-definition for the correct definition of left-hand derivative. Further, you seem to be just plugging in $a+h$ for $h$ in the first definition, so where does the $f(a+h-h)$ in the numerator come from? And also the $-h$ in the denominator? – K.Reeves Nov 30 '17 at 11:01
  • Sorry, I meant you seem to be plugging in $a+h$ for $a$ in the first definition. – K.Reeves Nov 30 '17 at 11:08
  • @K.Reeves I think that the left hand derivative at point $c$ is: $\lim_{x\rightarrow c^-}\frac{f(x)-f(c)}{x-c}$. Since x approaces from the left, we substitute $x=c-h$ to get $\lim_{h\rightarrow 0}\frac{f(c-h)-f(c)}{-h}$. To get the derivative at $a+h$, i substituted $c=a+h$ to get $\lim_{h\rightarrow 0}\frac{f(a+h-h)-f(a+h)}{-h}$ – Ryder Rude Nov 30 '17 at 11:11
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    If you are concerned with derivative at $a$ then why do you wish to deal with derivative at $a+h$? And even if you wish to deal with derivative at $a+h$ your handing for this is incorrect. – Paramanand Singh Nov 30 '17 at 11:23
  • @K.Reeves In the link you provided the definition is: $\lim_{h\rightarrow 0^-}\frac{f(a+h)-f(a)}{h}$. In this, we can substitute $h=-h$ and replace $h\rightarrow 0^-$ with $h\rightarrow 0^+$ to get the formula I used. – Ryder Rude Nov 30 '17 at 11:24
  • @ParamanandSingh I'm concerned with the left hand derivative at $a+h$ where $h$ is a positive infinitesimal. I've shown that this equals right hand derivative at $a$. And, how is my handling of this incorrect? – Ryder Rude Nov 30 '17 at 11:26
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    When $h$ is a positive infinitesimal then whatever you mention simply does not apply. Derivatives using infinitesimals are handled in a completely different manner than what you have written. The approach you mention assumes that $h$ is a real number and not an infinitesimal. Don't try to mix the infinitesimal approach with the usual real variable approach. – Paramanand Singh Nov 30 '17 at 11:28
  • You should have a look at the definition based on infinitesimal in this wikipedia article. – Paramanand Singh Nov 30 '17 at 11:32
  • What does "The LHD at the point $a+h (h\to 0)$" mean?.... "$The$ point"... Exactly which point ? – DanielWainfleet Nov 30 '17 at 18:18
  • If you work in an extension $\Bbb R^*$ of $\Bbb R$ with infinitesimals then for a function such as $f(x)=x^2$ the value of $|f'(a+h)-f'(a)|$ is smaller than any positive real when $h$ is infinitesimal but it is not $0.$ – DanielWainfleet Nov 30 '17 at 18:29

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