Derivative formulas through geometry | Chapter 3, Essence of calculus

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Strong rec Mateo_13 · 2022-05 · Mathematics Stack Exchange · on “Geometric Intuition for $\int^b_0 x^p dx= \frac{b^{p+1}}{p+1}$”

3 Blue 1 Brown has made an excellent video on a geometric way to understand the derivative of $x^2$ and $x^3$ : [this video]

Recommends Armin Meisterhirn · 2017-12 · Mathematics Stack Exchange · on “Derivative of square root”

Another possibility to find the derivative of $f(x)=\sqrt x$ is to use geometry. Imagine a square with side length $\sqrt x$ . Then the area of the square is $x$ . Now, let's extend the square on both sides by a small amount, $d\sqrt x$ . The new area added to the square is: $$dx=d\sqrt x \times \sqrt x + d\sqrt x \times \sqrt x + d\sqrt x^2.$$ [LINK] This is the sum of the sub-areas added on each side of the square (the orange areas in the picture above). The last term in the equation above is very small and can be neglected. Thus: $$dx=2\times d\sqrt x \times \sqrt x$$ $$\frac{dx}{d\sqrt x}=2 \times \sqrt x$$ $$\frac{d\sqrt x}{dx}=\frac{1}{2\times\sqrt x}$$ (To go from the second step to t

Recommends user10478 · 2018-06 · Mathematics Stack Exchange · on “Derivative of Square Root Visual”

Derivative formulas through geometry | Essence of calculus, chapter 3 ( 3Blue1Brown [LINK] ): [this video] There is a challenge at 12:23 [this video] asking the viewer to arrive at the formula for $\frac{d}{dx}\sqrt{x}$ by considering small changes in the length and area of a square. What is the correct way to approach this? I tried to copy the technique used to find $\frac{d}{dx}(x^2)$ at 2:25 by letting $u = \sqrt{x}$ . This makes the area added to the square $2u \, du + du^2$ , which comes out to $$2\sqrt{x}(d\sqrt{x}) + (d\sqrt{x})^2.$$ The second term is negligible, but how this equals $\frac12x^{-1/2}$ is still unclear.

Recommends Corvinus · 2018-09 · Mathematics Stack Exchange · on “Optimization explained to a middle school kid”

I very like the way you pose the problem and the visualization of it. Your last comment on applying similar triangle is a very good idea! It reminds me of an excellent video of 3blue1brown ( [this video] ) where he does a similar construction in order to show what the derivative $sin(\theta)$ is. Brilliant!

Recommends Nathan Bradley · 2020-01 · Mathematics Stack Exchange · on “Surface Area of a Hypercube”

The answer is $$2 \frac{d}{dx}(V_{n}(L))$$ or, equivalently: $$2nx^{n-1}$$ where $V_{n+1}(L)$ is the Volume of an n-dimensional sphere. As to why, think about slightly increasing the size of a cube for example. How much new volume has been created? it's going to approach 3x $^2$ times the tiny change in x, because each face of the cube will be expanded, and all the other bits will converge to zero. It's hard to explain, but here's a good video to watch to get a feel for what I'm talking about. [this video] However, this only covers half of what the surface area of these shapes is, because it's only increasing with respect to half of the total sides. So, multiplying by two will give your final ans

Recommends Abdulmalek Alem · 2022-06 · Mathematics Stack Exchange · on “Derivative of $ \frac{1}{x}$ geometrically”

Context: [LINK] "Essence of calculus" series called "Derivative formulas through geometry"- 3rd episode of chapter 3 , 3Blue1Brown [this video] : From 10:08 - 12:35 [this video] he discusses the derivative of $\frac{1}{x}$ geometrically. After the discussion of the basic idea, it is given as an exercise to solve. I tried solving it but I always come to conclusion that $\frac{d}{dx} \frac{1}{x} = -\frac{1}{x(x+dx)}$ , or $ -\frac{1}{(x^2 +xdx)}$ . Is it allowed to neglect " $x dx$ "? Or is there another correct method? My try: $$ \begin{align}A_{(lost)} &= A_{(gained)} \\ A_{(lost)} &= x * -d(\frac{1}{x}) \\ A_{(gained)}&= dx \frac{1}{(x+dx)} \end{align}$$ note:- $d(\frac{1}{x})$ is negative because th

Recommends Sammy Black · 2022-06 · Mathematics Stack Exchange · on “Why is derivative of $x^2$ not $x$?”

I recommend this video [this video] (actually the entire playlist) to help with *intuition* about the derivative. It clearly shows why $\frac{d}{dx} \bigl( x^2 \bigr) = 2x$.

Cites Yogi · 2018-08 · Mathematics Stack Exchange · on “Why is the derivative of $x^2$ equal to $2x$ rather than $4x$?”

Between 2m26s and 4m36s of this [this video] , the teacher explains the power rule using the area of a square. But why isnt the square or cube increasing its area or volume on all sides? If square increases its area by all 4sides, shouldnt the derivative be 4x dx ?? whats the argument behind this??? Thanks.

Cites user3180 · 2018-09 · Mathematics Stack Exchange · on “Show similar triangles on unit circle when proving the derivative of $\sin\theta$ geometrically”

3blue1brown attempts to prove the derivative of $\sin(\theta)$ using geometry at minute 16 of this video: [this video] [LINK] He says that these two triangles are similar. How do you show that?

Cites Guy Tabak · 2020-03 · Mathematics Stack Exchange · on “Derivative of $\sqrt{x}$ using geometry”

I am having trouble with a problem given in a video by 3Blue1Brown, to which I have already found a response, here [LINK] . My is issue is understanding why the equation $$\mathrm dx = 2(\sqrt{x})\left(\mathrm d\sqrt{x}\right)+\left(\mathrm d\sqrt{x}\right)\left(\mathrm d\sqrt{x}\right) = 2(\sqrt{x})\left(\mathrm d\sqrt{x}\right)+\left(\mathrm d\sqrt{x}\right)^2$$ was assembled (in the above thread). In the video, he does say dx = "New area" and by calculating "New area" I did reach to the same equation. My lack of understanding is why dx is regarded as the new area. Link [this video] to the problem in the video. Thanks :)