1

I'm reading through the proof of chain rule in Calculus with Analytical Geometry by George F. Simmons(Second Edition). The "almost" correct proof goes as follows. Here, y is a differentiable function of u and u is a differentiable function of x.

$$ \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta u} .\frac{\Delta u}{\Delta x} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta u} . \lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} = \frac{dy}{du} . \frac{du}{dx}$$

The text goes on to say that the above is invalid because it doesn't consider the case when $\Delta u$ is unchanged for a small positive change in $\Delta x$. I can see that if $u(x)$ is a constant function, this can happen, but we say in the theorem (rule) that $u$ is a function of $x$. In what other cases can this happen? It seems to me that if $u$ is differentiable, then it's also continuous. And for a continuous function that's not a constant, how can $\Delta u$ be 0, no matter how close $\Delta x$ approaches 0?

In the Wikipedia article, I saw an example of $u = x^2.\sin(\frac{1}{x})$, but $\sin(\frac{1}{x})$ is not continuous at $x = 0$.

Neo M Hacker
  • 141
  • 3
    $\sin(1/x)$ isn't continuous at $x=0$, but $u(x)=x^2\sin(1/x)$ is, and you can choose $\Delta x$ arbitrarily close to zero where $u(\Delta x)=u(0)=0$. I recommend graphing the function $x^2\sin(1/x)$ so you can see what's going on there - essentially, the zeroes of $u$ accumulate as you get nearer and nearer to the origin, but $u$ goes to zero fast enough that it's not just continuous but differentiable. – Steven Stadnicki Jun 01 '18 at 15:38
  • @StevenStadnicki Thanks for the answer. I should have tried to see that $x^2 \sin(\frac{1}{x})$ is continuous. I didn't know how to and it seemed like it wasn't.

    The plot looks insane: https://goo.gl/ABRsNk

    How would one go about proving that $x^2 \sin(\frac{1}{x})$ is continuous? I can see that if the factors individually have limits, then the limit of the product is equal to the product of the limits, but in this case limit of $\sin(\frac{1}{x})$ doesn't exist. So, that simple proof technique won't work. Do you know?

     – Neo M Hacker Jun 01 '18 at 15:52
  • Ah, obviously, it's been asked before: https://math.stackexchange.com/questions/232672/show-that-the-function-gx-x2-sin-frac1x-g0-0-is-everywhere?utm_medium=organic&utm_source=google_rich_qa&utm_campaign=google_rich_qa – Neo M Hacker Jun 01 '18 at 16:02
  • @StevenStadnicki So, a minor point. To be precise, $$ y = x^2 sin (1/x), x != 0 $$ $$ y = 0, x = 0$$ is continuous and the derivative of the above function is not continuous at 0. I also think as x approaches 0, the derivative doesn’t have a limit; the function is not differentiable at x = 0. – Neo M Hacker Sep 14 '18 at 16:59
  • The function $y$ in your last comment is differentiable at $x=0$ with derivative equal to $0$ but the derivative $dy/dx$ is not continuous at $x=0$. – Paramanand Singh Jan 18 '19 at 23:51

0 Answers0