Implicit Differentiation
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- Опубліковано 8 лют 2025
- This calculus video tutorial provides a basic introduction into implicit differentiation. it explains how to find dy/dx and evaluate it at a point. It also explains how to find the slope of a curve given the x and y coordinates. In addition, it explains how to find d^2y/dx^2, the second derivative of the function using implicit differentiation. You need to be familiar with the product rule, chain rule, and quotient rule for derivatives.
Derivatives - Free Formula Sheet:
bit.ly/4dThzf1
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Derivatives - Fast Review:
• Calculus 1 - Derivatives
Derivatives - The Product Rule - f*g:
• Product Rule For Deriv...
Derivatives - The Quotient Rule:
• Quotient Rule For Deri...
Derivatives - The Chain Rule:
• Chain Rule For Finding...
Derivatives - Composite Functions:
• Derivatives of Composi...
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Implicit Differentiation:
• Implicit Differentiation
Derivatives of Inverse Trig Functions:
• Derivatives of Inverse...
Derivatives of Exponential Functions:
• Derivatives of Exponen...
Derivatives of Logarithmic Functions:
• Derivative of Logarith...
Logarithmic Differentiation:
• Introduction to Logari...
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Derivatives - Using Logarithms:
• Finding Derivatives Us...
Derivatives of Inverse Functions:
• Derivatives of Inverse...
Derivatives - Differentiation Rules:
• Basic Differentiation ...
Derivatives - Function Notations:
• dy/dx, d/dx, and dy/dt...
Derivatives - The Reciprocal Rule:
• The Reciprocal Rule an...
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MR. Organic Chemistry Tutor, thank you for an excellent video/lecture on the classical Implicit Differentiation in Calculus I/II. This is an important topic in Mathematics and Engineering. This is an error free video/lecture on UA-cam TV with the Organic Chemistry Tutor.
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It's so gooood, realizing that d/dx y^3 = d/dx * y^3/dy * dy = d/dx * 3y^2 * dy = 3y^2 dy/dx made my day
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Shortcut for implicit differentiation
Take all the things on one side
then do as below:
dy/dx= -(diffentiate wrt x, take y as a constant)/(differentiate wrt y, take x as a constant)
Eg. Lets say I need to differentiate x³+3xy=y²+1
Step1
(taking everything on one side)
x³+3xy-y²-1=0
Step2
differentiate x³+3xy-y²-1 wrt x and take y as a constant
That would be equal to 3x²+3y
Let 3x²+3y = m
Step3
differentiate x³+3xy-y²-1 wrt y and take x as a constant
That would be equal to 3x-2y
Let 3x -2y = n
Step4
Apply the rule and obtain dy/dx
dy/dx = -m/n = -3x²-3y/3x-2y
This may look like this is quite lengthy but in actual it hardly takes 10secs. I wrote this long so that anyone who sees this could clearly make sense of it and understand
Interesting method, thanks.
Thank you sir may I ask you a question
7:23 how did -2x=cos xy² become y²cos xy² ??? Where did y² came from ???
@@Maroofgilani621 he distributed cos(xy^2)
@@Maroofgilani621 Chain rule : d/dx sin(u) where u = xy^2
d/dx (sinu) = d/du (sin u). du/dx
d/du (sin u) = cosu
du/dx = y^2 + x.2y.dy/dx
Then cos(xy^2) . (y^2 + x.2y.dy/dx)
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This gave me the key points needed to proceed with implicit differentiation.
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13:35 I believe you could have substituted 9 in for (y^3 + x^3) as well, thank you for the help!
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gah i am so confused why we can use dy/dx = -x^2/y^2 in the quotient rule result but not for an immediate implicit differentiation of 3x^2 +3y^2(dy/dx) -----> 6x+6y(dy/dx)+3y^2(d2y/dx2)(dy/dx). it obviously doesn't seem to math out to the same answer....just feels weird that this doesn't also work
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12:56 To add, notice that y³ + x³ is also equal to our original given/expression x³ + y³, and it states that it is equal to 9. So subtituting the value of that expression will yield to a same answer. This is the beauty of second derivate in implicit, original expression tends to have a relationship on some part of it.
I have a question,so you mean second derivative will always give you the answer of the original equation if you substitute???
@@nokuthulamakhunga7039 Actually, there are times that a given will not give you coordinates like (1,2) etc. But always remember that in implicit differentiation - second derivative, there's a higher chance that your solution will consist of the original given equation like for example here was x³ + y³ = 9, only the majority though, because some are not. Sometimes, you have to make little algebra tricks to make show of the original expression. For example, instead of getting y³ + x³, you get -y³ - x³, what you're going to here is to multiple it by "-" or -1, so it will become -(y³+x³) which will be equal to -(9) = -9.
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0:01: ✏ Implicit differentiation is a method used to find the derivative of an equation with respect to x, by treating y as a function of x and adding dy/dx to every term containing y.
4:31: 💡 The video explains how to use implicit differentiation to find the derivative of a function and evaluate it at a specific point.
10:55: 📈 The video discusses finding the second derivative of a function using the quotient rule and evaluating it at a specific point.
Recap by Tammy AI
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I was totally following until you changed from dividing to distributing at abouit 6:48. I didn't track how the distribution worked.
particularly when you distributed it to the derivated of xy^2. I don't see how the x ended up in the derivative of 1y^2, instead of it being 2y, or what happened to the 2y where it's x(2y(dy/dx))
Does anyone understand why you add the (dy/dx) and how to know when to add it? I'm confused, how do you know when to put the extra differentiation
You add it when differentiation y variables
You put it on every variable except x
Because dx/dx cancels itself out
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at 8:24 , could we cross cos(xy^2) in both numerator and denominator
@toshio3772forgot about that, thanks for clearing it up
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1:00 You don't really add the DY/DX. You times. But including it is a figure of speech. So it's a figurative add. But it will be taken literal and confuse users.
So I don't have to write it out on paper like that?
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As I understood it now, implicit differentiation is simply derivation with extra precision to what we are differentiating in respect of what. dy/dx. dr/dt. dv/dt ...etc. But normally d/dx is d to respect of X assumed and does not follow the rest of calculation.
Yeah I think it attempts to relate y and x even though they’re visibly (in the equation) not in the regular y = x form. I keep hearing that it applies the chain rule, but if that were the case, then we’d need to treat y not as a single variable but as y(x) or a function of x. (Chain rule in y(x) 👉 would be y’(x) • x’.)
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at 2:14 would'nt it be -8/6 because it is dy/dx ?
Why in the living %^*# is there a negative placed there though
It's -x/y
The thumbnail scared me with the -x^2/y^2 i thought everything i knew about derivatives was wrong lol
8:22 , is it considered when you simplify the final answer by expanding it.?
at 7:36, can you explain how you got 2xycos(xy^2)dx/dy?
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7:23 how did -2x=cos xy² become y²cos xy² ??? Where did y² came from ???
He multiples the cos(xy2) into the 1y squared. He also multiples it into the other term that's being added in the parenthesis.
thanks for this lesson
How do you know which terms have the dy/dx?
whenever you differentiate y
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