2

If the derivative of $f(x)$ didn't exist at $x=a$. Then it automatically means that its left-hand and right-hand derivative are not equal.

My question to know differentiability of $f(x)$, do we need to evaluate left and right hand derivative. If not, when do we require them?

Thanks.

Bernard
  • 175,478
Rajesh Marndi
  • 469
  • By definition, yes! You need to evaluate both lateral derivatives. But there's some conditions you could check to ensure the existence (or not) of derivative. For instance, if the function isn't continuous on a point $x=a$ then the derivative doesn't exist on that point. Or, if a function is a polynomial function, you could conclude that it has derivative in $\mathbb{R}$. – Pspl Aug 29 '19 at 12:50
  • If the function isn't continuous at x=a, will we get a value if we derivative at that point. I believe derivative will not exist at that point. – Rajesh Marndi Aug 29 '19 at 13:00
  • I hope you are talking about derivative in $\mathbb{R}$, because in $\mathbb{R}^{n}$ for $n \ge 2$, left and right derivatives does not make sense. – Manoj Kumar Aug 29 '19 at 13:05
  • No, I am talking about 2-dimension only.So if f(x) isn't continuous at x=a, can derivative exist at x=a? – Rajesh Marndi Aug 29 '19 at 13:28
  • A differentiable function is continuous, so if the function is not continuous at a point, it cannot be differentiable there. Also there are many cases where we know the derivative exists: sum, product, composition of differentiable functions, for example. Then we don't need to compute both one-sided derivatives, because we know in advance that they are equal. – saulspatz Aug 29 '19 at 13:45
  • Can I have some example where both one-sided derivative are required to tell f(x) differentiability at x=a, when f(x) is continuous at x=a. – Rajesh Marndi Aug 29 '19 at 13:54
  • The trivial example $f(x)=|x|$ will show you that, although $f$ is continuous at $x=0$, the derivative doesn't exist (because $f'(0^-)=-1\neq 1=f'(0^+)$). Usually this happens when the function line makes a «corner» or the tangent line to it is vertical. – Pspl Aug 29 '19 at 14:36
  • That is my question, why do I require to do both sided derivative to determine its differentiablity. If I derivative f(x)=|x| , It will give x/|x| which shows it doesn't exist at x=0. – Rajesh Marndi Aug 29 '19 at 14:43
  • 1
    The general rule is "use the definition to check the differentiability on a changing sub-function point on a piecewise function" - since an absolute function is also a piecewise, it makes sense to use the very same definition. That's because all elementary operations, preserves differentiability. For instance, to justify the differentiability of $y=\sin x$ all you have to do is state that is a trigonometric function, so it's differentiable. – Pspl Aug 29 '19 at 14:52
  • 1
    If $f(x) = |x|$, then the expression $f'(x) = \frac x{|x|}$ is only true for $x \ne 0$. It doesn't itself tell you anything about the existence of $f'(0)$ ($\frac x{|x|}$ is undefined at $0$, but you are not guaranteed that $f'(0)$ is the same). To determine that, you need to look at the rh and lh derivatives at $0$. – Paul Sinclair Aug 30 '19 at 02:41

0 Answers0