Implicit Differentiation - Find The First & Second Derivatives
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- Опубліковано 5 лют 2025
- This calculus video tutorial provides a basic introduction into implicit differentiation. it explains how to find the first derivative dy/dx using the power rule & product rule. It also explains how to find the second derivative d2y/dx2 using the chain rule.
Derivatives - Free Formula Sheet:
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Derivatives - Free Formula Sheet: bit.ly/4dThzf1
Derivatives - Video Lessons: www.video-tutor.net/derivatives.html
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Second derivative starts at 7:20
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@RandomGuy Most likely his initials.
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Why are we multiplying by y at 10:40?
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very good examples of implicit differentiation
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Hi, I managed to understand everything so well. I do have one question though, why do we multiply most of the derivatives that we get from power rule/product rule by dy/dx? Whats the significance of dy/dx and where does it come from? Thank you!
Dy Dx is the two sides of the triangle connected to the tangent line when your trying to measure an implicit curve. To visualize it think of a unit circle centered at 5x and 5y. His initial equation is the implicate curve and the dirivitive will give you the slope of the tangent line.
If I'm saying this correctly you use implicit differentiation when your finding the tangent line only when you have a graph that would look like a complete circle? The channel 1brown3blue really explains the deeper theory well with lots of visuals.
Just like when you learn to do integration which is measuring the area under the curve dx is the space of the many rectangles your measuring the area of it it means differentiate x in terms of derivative of y I believe... I may have the last part not 100 percent. Got t go look that up again.
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Why did you multiple y on both denominator and numerator on the second derivative
At 10.30 minutes
can i ask why do we write dx² and not d²x ?
When I took calc in high school, this chapter test only had one second derivative problem because it’s so long and tedious with multiple substitutions. Now I’m in college and our test had 3 second derivative questions, like wtf there’s so much room for errrrrrror. Damn you college
Unless dy/dx = x/y or y/x then these second derivatives kick my ass
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Wow! This is so helpful
Which software do you use for your tutorials (not for screen recording but for illustrations). Thanks.
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(Pertaining to the last example) Why do you have to implicitly derive the first time but then derive normally the 2nd time? I understand that they're both implicit differentiation, but I dont understand why you cant just solve for y the first time or derive it without solving for y' the 2nd time.
both are implicitly derived
I prefer you way more than Khan academy
10:45 Why didn't you multiply x^2 by Y as well in the numerator?
(x * -x/y) * y will turn out to be (-x^2/y) * y
when u evaluate that it will turn into (-x^2 * y) / y
cancel out the y and it becomes -x^2
he just skipped that step thats why it was confusing
At 2:03 ,why is dy/dx written with 3y^2?
bc you're differentiating in terms of x, so when you take the derivative of a y variable, you need to multiply it by dy/dx to make it in terms of x
@@fluffyemu2714 Thank u very much.I got it.
I tried finding the 2nd derivative of the 2nd problem. I stopped halfway since it was taking too long. Is it really supposed to take too long, or am I not doing it right?
Quotient rule is suppose to use subtraction in numerator right?
9:38 why is (dy/dx)=(-x/y)
Why you written dy/dx with 2y can you explain it
Very helpful 😊
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Why did dy/dx come after 3y^2 instead of coming after -4 or after 2x
why do you keep dy/dx? I am confused like where does that come from
Well in finding the second derivative of x square + ysquare= 25. In. -y+x(-x/y)/ysauare. Y r u multiplying the top and bottom by y.that's where I'm confused😢
In the second line he multiplies 3y^2 by (d/y divided by d/x), does that signify that the slope of 3y^2 is different from that of y^3 and the d/y divided by d/X adjusts for that?
In 11:26 i think -3^3 is 9 not -9 is it right ?
First, you take a square of (3)^2, and then the answer 9 goes with the minus sign.
Solve the parenthesis first because minus is for the answer of the parenthesis.
Pls upload a video of hyperbolic differentiation. Thanks 😊
please tell me, why the derivative of y is equal to -x/y?
thank youu so much for this!
May I ask why you multiplied it by Y at the end part?
if you were to solve the numerator, you would have (-y^2 - x^2) / y. When you have a fraction inside of a fraction, you kiss and flip. i was confused too but if you're really having touble just solve that equation and you'll be like "ohh!"
Thanks you so much sir
Very helpful ❤️🤠
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How does x time y become x squared
Woah waoh am i hearing this correctly? 0:02
Paddletatoe 77 Oml I was thinking the same thing. 👀
11:14 where did you plug from???????????? Where did ... oh ok I get it thanks
Is there a way I could use this d^2y/dx^2 within a point slope formula or something similar to form tangent lines through a point given that isn't present in the initial function? I've been doing this in a simple form with parabolas and other basic curved lines that use only d/dx. I use point slope formula with "d/dx of original function=(given y point not in function minus f(x))/(given x point not in function minus x)" to find the possible inputs of x in the original function then work from there. This equation worked nicely because I had d/dx = an equation with only x variables and constants. However, I'm definitely struggling to do the same thing with hyperbolas, circles and ellipses.
.
An example of what I'm asking would be. Find the equation of the tangent lines of the equation 1=x^2/9+y^2/25 that pass through the point (7,-2). Might this utilize d^2y/dx^2?
Why multiply by Y again?
Did you find answer? Why?
@@diana-lb6fu if you were to solve the numerator, you would have (-y^2 - x^2) / y. When you have a fraction inside of a fraction, you kiss and flip. i was confused too but if you're really having trouble just solve that equation and you'll be like "ohh!" its just basic math but he skipped a step which made me confused too
You are amazing
Can the first derivative shown in the video also be solved as
y'=1/3(-x^2+4x+8)^(-2/3)*(-2x+4) after we isolate y^3 and take the cube root of both sides?
No it’s wrong
THANKS
thank u😭
what if I'm told to find d/dy??? I cant find anything that teaches this
I dont get why is v’ equal dy/dx shouldn’t it just be 1 because The derivative of y equals 1

God bless you!
can I ask where did he get the P(4,3) for finding the second derivative?
Given