Is the Derivative of a Differentiable Logistic Function Continuous
Why is/isn't the derivative of a differentiable function continuous?
Solution 1
The theorem is simply stating that the function \begin{align*} \varphi(x) &= \begin{cases} \frac{f(x) - f(a)}{x - a} & \text{if}\;x\not = a; \\ f'(a) & \text{if}\;x = a \end{cases} \end{align*} is continuous. And it clearly is; the only point to check is $x = a$, and the condition $\lim_{x\to a} \varphi(x) = \varphi(a)$ is exactly the definition of $f'(a)$. The theorem is not claiming that $f = \varphi$ everywhere on $I$.
One of the classic examples of a differentiable function $f$ with $f'$ not continuous is $f(x) = x^2\sin (1/x)$ (with $f(x) = 0$). The derivative \begin{align*} f'(x) &= \begin{cases} 2x \sin (1/x) - \cos (1/x) & \text{if}\; x \not = 0; \\ 0 & \text{if}\; x = 0 \end{cases} \end{align*} exists everywhere, but it is not continuous at $0$.
Solution 2
The point is that $\varphi$ is not $f'$. They just coincide in one point, and it is easy to see that two functions coinciding in one point entails nothing about some relationship of differentiability/continuity etc between one another.
Solution 3
The theorem states that $\varphi(a)=f'(a)$ for this particular value of $a$. It doesn't say that $\varphi(x)=f'(x)$ for all $x$, or indeed for any value of $x$ besides the single value $x=a$. So the fact that $\varphi$ is continuous at $a$ doesn't tell you that $f'$ is continuous at $a$, since continuity depends on the values of the function at points near $a$, not just at $a$ itself.
Related videos on Youtube
32 : 48
Continuity and Differentiability
11 : 39
Proof: Differentiability implies continuity | Derivative rules | AP Calculus AB | Khan Academy
09 : 38
Differentiability and continuity | Derivatives introduction | AP Calculus AB | Khan Academy
09 : 11
Continuity vs Partial Derivatives vs Differentiability | My Favorite Multivariable Function
06 : 34
3.9 Continuous but not differentiable functions
Comments
-
I am confused about the following Theorem:
Let $f: I \to \mathbb{R}^n$, $a \in I$. Then the function $f$ is differentiable at $a$ if and only if there exists a function $\varphi: I \to \mathbb{R}^n$ that is continuous in $a$, and such that $f(x) - f(a) = (x - a)\varphi(x)$, for all $x \in I$; furthermore, $\varphi(a) = f'(a)$.
I understand the proof of this theorem, but something confuses me. Doesn't this theorem state that the derivative of a function in a point is always continuous in that point, since $f'(a) = \varphi(a)$ is continuous in $a$? This would mean that the derivative of a function is always continuous on the domain of the function, but I have encountered counterexamples. I have probably misinterpreted something; any help would be welcome.
-
You mean $\varphi$ is continuous at $a$.
-
What does I mean in this question? Is it a continuous interval of R? Must it be open?
-
-
"the only point to check is $x=a$": Actually, the theorem in the question only states that $\varphi$ is continuous in $a$ (and does not assume $f$ to be continuous).
-
Makes sense if $f$ is only assumed to be differentiable at $a$ (rather than everywhere), but I initially interpreted "continuous in $a$" as a typo for "continuous in $x$" rather than a typo for "continuous at $a$".
-
This goes further: there is no reason for $f'$ to even exist.
-
The "if and only if" would be false if continuity anywhere other than at $a$ would be required. A counterexample is easily provided by a function $f$ that is differentiable at $a$ and discontinuous at some $b\ne a$.
-
What are open/closed intervals?
-
@Stefan I believe you are correct. Thankfully the main point of this post is correct even with that error in place. And since this is SO, I don't even know if I'm welcome to edit the post to correct the error.
Recents
Related
Source: https://9to5science.com/why-is-isn-39-t-the-derivative-of-a-differentiable-function-continuous
0 Response to "Is the Derivative of a Differentiable Logistic Function Continuous"
Post a Comment