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Right from 10th, my teachers gave me all the formulas and techniques on how to solve problems in calculus.

Product rule= $$\frac{d}{dx}f(x)g(x)= f(x)g'(x)+g(x)f'(x)$$

Chain rule= $$\frac{dy}{dx}=\frac{dy}{du}* \frac{du}{dx}$$

Why is it that while applying product rule, one function is kept constant, while the other differentiated?

I saw the video on the visual representation of the chain rule and product rule on 3Blue1Brown, yet I'm unable to formulate an explanation for it. Especially when we increase the number of functions and variables.

I would really appreciate any help regarding this. Thanks in advance!!

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    $\begingroup$ That is not the chain rule, you're referring to the product or Leibniz rule. $\endgroup$
    – Javi
    Dec 29, 2020 at 13:27
  • $\begingroup$ You might want to have a look at a proof of the product rule, even if this is not a fully conceptual explanation, proofs usually give "reasons" for why something is true. $\endgroup$
    – Javi
    Dec 29, 2020 at 13:31
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    $\begingroup$ I'm sorry, I have made the correction...I was fussing over the problem for too long and confused myself $\endgroup$
    – Smriti Sivakumar
    Dec 29, 2020 at 13:31
  • $\begingroup$ Thank you, I did go through this. As you rightly mentioned, the explanation isn't conceptual, which is why I posted the question here $\endgroup$
    – Smriti Sivakumar
    Dec 29, 2020 at 13:33
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    $\begingroup$ "An Intuitive Proof of The Product Rule" with diagrams, and an extension question into 3 dimensions in the follow on exercise (Question 5) PDF : numberwonder.co.uk/Download/PD9028/9028(4).pdf $\endgroup$
    – Martin Hansen
    Dec 29, 2020 at 16:49

1 Answer 1

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I like to think about how the area of a rectangle changes when we vary the sides:

$$(a+\delta_a)(b+\delta_b)-ab=\delta_a b+a\delta_b+\text{ something too small to worry about.}$$

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  • $\begingroup$ Based on this, if we have n functions, can we say we're finding the volume of the object in an n-space? Please correct me if I'm wrong... $\endgroup$
    – Smriti Sivakumar
    Dec 29, 2020 at 13:48
  • $\begingroup$ Well, if you like. But this is only to help our intuition.. $\endgroup$
    – ancient mathematician
    Dec 29, 2020 at 13:49

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