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I looked at the geometric meaning of derivatives and the chain rule video in the 3blue1brown calculus series.

https://youtu.be/YG15m2VwSjA

My question is that when we calculate the derivative of the product of two functions, we disregard the little rectangle in the bottom right. Please see the chain rule video at 6;20. Suppose we have a function where that quantity is so huge that we cannot disregard it.

How do we understand product rule geometrically using the methodology in that video in this case?

I guess I’m so confused that I cannot even formulate the questions properly :-)

thanks, GK

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    $\begingroup$ If a function is differentiable, then by by definition we can zoom in so closely that it is basically linear. In other words, the function you allude to would not be differentiable in the first place. $\endgroup$
    – pancini
    Oct 31, 2019 at 19:02
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    $\begingroup$ As Elliot aptly notes, there are in fact functions where the quantity you refer to is too large to ignore. These functions are not differentiable. You can't apply any of these rules there. These rules say, that assuming that locally, everything looks linear, we can do these things. Great question BTW $\endgroup$
    – Rushabh Mehta
    Oct 31, 2019 at 19:05
  • $\begingroup$ @DonThousand: Can you give examples of a functions where the quantity is "too large to ignore"? By definition, the $dx$ differentiation can always be made infinitesimally small so the product terms are negligible.... no? $\endgroup$
    – David G. Stork
    Oct 31, 2019 at 19:08
  • $\begingroup$ It goes too far to ask from us to view a video (incl. spam) for $6$ minutes. $\endgroup$
    – Christian Blatter
    Oct 31, 2019 at 19:13
  • $\begingroup$ 2@DonThousand; thanks.makes thing clearer. Please write as an answer so I can upvote. $\endgroup$
    – Khanna111
    Oct 31, 2019 at 19:19

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I'm usually not a fan of how derivatives are defined and are explained from an intuition perspective, since the focus is usually on how to take the derivative, rather than why we can take the derivative.

Note that the derivative of a single variate function $f(x)$ at $x_0$ is the limit $$\lim\limits_{h\to0}\frac{f(x+h)-f(x)}h$$The existence of this limit roughly implies that there exists some tangent line at $x_0$ to $f(x)$ such that the function will be arbitrarily close to the tangent line depending on how far I zoom in. (I'm not usually a fan of thinking about derivatives in terms of tangents, but I think that in this context, it is appropriate).

Are there functions where this approximation fails miserably? Most certainly. In fact, most continuous functions are not differentiable (and most functions are not continuous). One of the first continuous functions that was discovered to be nowhere differentiable was the pathological Weierstrass Function.

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