finding dy/dx of a heart curve by using implicit differentiation, calculus 1 tutorial

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Derivative of a heart curve by implicit differentiation. This is definitely one of my favorite curves in calculus and we will find dy/dx by using implicit differentiation. If you are taking calculus 1, then this is a good practice to master your derivative skills.

#calculus #blackpenredpen
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derivative of a heart should be 0 as love should be constant <3

peakpeet
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"Computer programmers; they have to be extremely patient because the codes are just crazy." - Amen #yay

alexdarcovich
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Hey bprp i got a better answer .


At the start just add x^2*y^3 to both sides and then take the cube root on both sides and then perform implicit differentiation

The answer by this method is

(2y-6x^(4/3) )/[3x^(1/3)][2y-x^(2/3]

hamiltonianpathondodecahed
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I did this derivative back in my Calculus I class and arrived at the same answer, but if you notice, there is a problem with the answer that you'll quickly find if you evaluate dy/dx at x = -1 or x = 1. At the points (-1, 0) and (1, 0) the slope is indeterminate despite the fact you can see on the curve that it has a definite slope. The numerator and denominator both evaluate to 0 with both points in it's current form.

bilbobaggins
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Title should read: Love is complicated, not anymore with math.

TheREtRoLoVeR
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Can you solve the original function for y in terms of x to have a nice function?

andi_tafel
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These videos are great for refreshing my old brain, young man. Thanks!

bigrobbyd.
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Solving this equation on Valentines Day will be An achievement

zat
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As pointed out below one can obtain a function for both top and bottom halves which is The bottom half (the -ve sqrt, below y=1/2) includes the point (1, 0). One can differentiate and find dy/dx=2 at x=1. The plot of the graph in the video shows that is reasonable also. But the implicit formula derived here gives dy/dx = 0/0 (=?) for the point (1, 0). What does this implicit result thus mean?

SBGif
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Now find the integral.









(Not really hard if you just solve for y first.)

andi_tafel
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Wow that looks like a complicated derivative. Try to solve that as a differential equation

helloitsme
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Substitute (x^2+y^2-1)^2=x^(4/3)y^2
dy/dx =

keithun
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I will be finally be able to derivate her heart :v #yay

pablojulianjimenezcano
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using partial derivatives fx, fy >>>> dy/dx=-fx/fy

قناةالاستاذأحمدالألفيالتعليمية
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I want to know what can we do with the knowledge of implicit differentiation, like how can we use it to achieve other stuff in calculus

shlomi
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One problem I had as a student was knowing when to stop. I would have spent a bunch of time trying to simplify the answer you reached. How do you know not to do that?

brianxx
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What is the equation and its derivative represented in standard form

its_robbietime
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My calculus teacher probably would have asked to clean up and simplify what you solved...and even graph it...find intercepts and find the double integral.

joeli
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whats the difference between d/dx and dy/dx ?

nigeltims
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What does the graph of the derivative look like?

aymanabdellatief