Visualizing the chain rule and product rule | Chapter 4, Essence of calculus

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Recommends WB-man · 2017-06 · Mathematics Stack Exchange · on “Why do derivatives behave like fractions when they are not fractions but an operator as a whole?”

3blue1brown has a video [this video] that may help with this. The link takes you to the relevant part of the video, but I recommend watching the whole thing if any part of it seems confusing.

Recommends rb612 · 2018-10 · Mathematics Stack Exchange · on “Chain rule as a product of slopes”

@xevepisis yes, you got it! As a silly but more concrete example, if me taking a step causes my dog to take 3 steps, and for every step my dog takes, my hamster take 2 steps, then every step I take will make my hamster take 6 steps! :) And I highly recommend 3blue1brown where he goes into more depth into this "nudging" intuition. See the video here [this video] .

Recommends Isaac Browne · 2020-04 · Mathematics Stack Exchange · on “Spivak Chapter 10, Problem 29 : why is this true?”

I believe that the problem here is a lack of intuition for the product rule. You apply it to derive $h'(0) = 0$ , but you seem to not see intuitively why it must be $0$ and not $1$ . 3Blue1Brown has a nice exposition of the general intuition behind the product rule (and the chain rule!) here [this video] . Thus, instead of trying to replicate that, I'll just speak about how to apply it here. Most people will show you the following picture for the product rule: [LINK] The strange thing about this case is that the part that matters to us is "Area C". This is because $x=0$ , so $f(x)=g(x)=0$ . Thus this will be the only nonzero area in the entire picture! Now, the other critical piece of intuition i

Recommends Joe · 2020-08 · Mathematics Stack Exchange · on “How would you calculate a derivative of $ f(x)= \frac{\sqrt{x+1}}{2-x}$ by the limit definition?”

For what it's worth, the quotient rule can be derived from the chain rule and product rules if you are familiar with those. If not, then I'd highly recommend watching this [this video] video from 3Blue1Brown (he makes maths intuitive):

Recommends Joe · 2020-10 · Mathematics Stack Exchange · on “Is it possible to turn this 'proof' of the product rule into a rigorous argument?”

I have often found linear approximation to be useful in understanding the main theorems of calculus. I tried using it to 'prove' the product rule, as I find the typical proof for it to be unintuitive. However, I'm not sure that the substitution I made can be properly justified: $$ (f \cdot g)'(a) = \lim_{h \to 0} \frac{f(a+h)g(a+h)-f(a)g(a)}{h} $$ Here is where I use my questionable substitution: replace $f(a+h)$ with $f(a)+f'(a)h$ ; make a similiar substitution for $g(a+h)$ . As $h$ approaches $0$ , the linear approximation becomes better and better. $(f \cdot g)'(a)$ becomes \begin{align} \lim_{h \to 0} \frac{\bigl(f(a)+f'(a)h\bigr)\bigl(g(a)+g'(a)h\bigr)-f(a)g(a)}{h} &= \lim_{h \to 0} \fr

Recommends rb3652 · 2021-06 · Mathematics Stack Exchange · on “Multivariable Chain Rule Proof”

I'm trying to understand the Proof of Chain Rule for functions of 1 independent variable and 2 intermediate variables. Here's my reasoning, step-by-step: The book reasons that the proof "consists of showing that if $x$ and $y$ are differentiable at $t=t_0$ , then $w$ is differentiable at $t_0$ , then $\frac{d w}{d t}\left(t_{0}\right)=\frac{\partial w}{\partial x}\left(P_{0}\right) \frac{d x}{d t}\left(t_{0}\right)+\frac{\partial w}{\partial y}\left(P_{0}\right) \frac{d y}{d t}\left(t_{0}\right)$ ". That makes exactly zero sense to me. Why do we have to start like this? Is there no geometric proof, akin to 3Blue1Brown's beautiful, intuitive picture of sliders [this video] for the independent, int

Cites Gabriel Alfour · 2018-07 · Mathematics Stack Exchange · on “Intuition behind neglecting higher order differentials in visual proofs of the Product Rule”

In most visual explanations of the product rule (like this video [this video] ), the justification for the remaining square ($df\cdot dg$) being neglected is that it is "very small". But the whole point of the intuition is that the $df \cdot g$ and $dg \cdot f$ are very small too. As such, I'm interested in either an intuitive explanation of this, or an other intuitive explanation for the product rule that takes this into account. Thanks in advance

Cites Khanna111 · 2019-10 · Mathematics Stack Exchange · on “Derivative when $f(x+dx)$ is hugely different from $f(x)$”

I looked at the geometric meaning of derivatives and the chain rule video [this video] in the 3blue1brown calculus series. 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 [this video] . 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 :-)

Cites Frank · 2021-01 · Mathematics Stack Exchange · on “Visual proof for $\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$”

I've found visualizations of simple mathematical concepts to be a really useful tool in building intuition for more complex mathematical concepts. For example, this visualization [LINK] of $(a + b)(c + d) = ac + ad + bc + bd$ can be used to visualize the calculus product rule [this video] . That being said, does anyone know of a visual way to show the following equation? $$ \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd} $$