Love this video so much! I was craving to find the video that show me derivative using transposition of terms, it makes me know why and when I should use the implicit differentiation clearly.
I like to teach chain rule outside first, then inside(s). In this order, the implicit is even easier - because you keep differentiating until you hit the bare 'y' term - and write dy/dx. Downside of reversing the order is you need to clean up the expression in most cases, but I think that's a benefit.
Thank you very much now i understand differentiation watching ur series of videos about it instead of just memorizing the rules of differentiation and applying it !
In Implicit functions (IF) the "Dependent variable (DV) is NOT JUST a function of the Independent vaiable (IV)". So if IV is X then DV might be Y or G or a mixture of X and Y but not JUST a function of the IV or X. Some use explicit = numerical quantities and implicit for non numerical quantities but when dealing with (IF) it is more comprehensively defined as "When the DV is not just a function of IV" Differentiation of that DV which is not a function of IV requires the chain rule.
Why do we have to use the chain rule when differentiating y²? Does the Power Rule not work? I'm thinking it might work, but we don't use it because that way we wouldn't have that dy/dx term, correct?
A circle can be expressed as 2 functions: y = sqrt(r^2-x^2) (top half) and y = - sqrt(r^2-x^2)(bottom half). when I do the derivitave of the top half I get -x/y, just like with implicit differentiation. But when I do the bottom half, I get positive x/y. What am I doing wrong?
You are doing nothing wrong: in the bottemleft quadrant x/y is negative. In the equation x (negative) is divided by y (positive because sqrt( r^2-x^2) stays positive). Example at (-3, -4) the equation gives (-3)/(sqrt(25-(-3)^2)= (-3)/(4). This is true since at the bottomleft quadrant the slope is negative.
at 10:19 why did you change (d/dx)y to dy/dx, because aren't you essentially going from the derivative of y wrt x to the rate of change of y to x which are two different things?
In this context,"gradient" means the actual numerical value at some specific point on the curve. The derivative is the formula to calculate the gradient. He already has the derivative. It is -x / y. Now he plugs in the values and gets the actual gradient at the point in question. Remember: "gradient" is just a fancy word for "slope of the tangent". But this is a numerical value. The derivative is a formula to calculate this slope at any point on the curve. The derivative describes in which way this slope changes when walking along the curve.
He made 2 errors. First the sq. root is +/- and not just + making the first method more tedious. Second one should show all the steps of the chain rule making the second method look easier and the preferable method since one CANNOT always make Y explicit. So when trying to find dY^2/dX by chain rule he should show all the steps. So initially we should write d Y^2/dY and not dY^2/dX. Then one can easily see that dY^2/dY = 2Y and dY^2 = 2Y*dY. Then dividing both sides with dX we get dY^2/dX = 2Y*dY/dX.
does anyone else answer his questions? xD like in the video when he asked if we were happy with how he got to the solution, i was like "wait *pauses* *scans the board* okay yea i got it". xD
9 years later and still being insanely helpful, what a treasure.
I was conceptually stuck differentiating y^2 with respect to x and this was the perfect explanation that I needed! Fabulous, thank you!
This helped me so much. Thank you Eddie, you don’t know how grateful I am for your work!
Great way of comparing the two ways of solving for dy/dx. Thanks.
This is the thing, you focused on what is the most confusing part of solving implicit differentiation problems and made me understand. thank you :)
You explain things so well! Thank you!!
thx for the explanation for why we multiply dy/dx when ever we differentiate y.
Love this video so much! I was craving to find the video that show me derivative using transposition of terms, it makes me know why and when I should use the implicit differentiation clearly.
The only vid I've found that really explained the extra dy/dx part.
THIS MAN IS THE GOATTTTTTTTT
An incredible explanation; it helped me understand very easily. Thank you so so so much!!!
Thank you so much! You are the only one explained why y² became 2y·y' after differentiation
I like to teach chain rule outside first, then inside(s). In this order, the implicit is even easier - because you keep differentiating until you hit the bare 'y' term - and write dy/dx.
Downside of reversing the order is you need to clean up the expression in most cases, but I think that's a benefit.
The only video that made me understand
Thank you very much now i understand differentiation watching ur series of videos about it instead of just memorizing the rules of differentiation and applying it !
I really don't know how I would pass calculus without you
TRULY A GOD TIER EDUCATOR
This explanation is fantastic!
I feel like i'm in class...
Great video on implicit differentiation. Thank you!
you explain things so well, thank you!
In Implicit functions (IF) the "Dependent variable (DV) is NOT JUST a function of the Independent vaiable (IV)". So if IV is X then DV might be Y or G or a mixture of X and Y but not JUST a function of the IV or X. Some use explicit = numerical quantities and implicit for non numerical quantities but when dealing with (IF) it is more comprehensively defined as "When the DV is not just a function of IV"
Differentiation of that DV which is not a function of IV requires the chain rule.
best math teacher ever
Brilliant explanation
Wtf i was i tuned this with headset on almost busted my eardrum
The passion is just pouring 7:30. Just a constant 😭😷🚀
Sir can you please make video on partial differentiation from basics.... Because it is creating problem for me in fluid mechanics
Cheer~~~implied though not plainly expressed.😊
man, I really felt bad when he had to pass the best part which he differentiates y. He thought that sudents couldnt understand him and I feel him.
Great ❤️❤️❤️💕
Excellent explanation
Great 🍺🍺🍻.
Thanks a lot for such easy explanation!
Interesting
Eddie Wooooooo 💫 💫💫
thank you man may Allah bless you
Why do we have to use the chain rule when differentiating y²? Does the Power Rule not work? I'm thinking it might work, but we don't use it because that way we wouldn't have that dy/dx term, correct?
If you use the power rule on y², which is y*y, you would get y*y' + y'*y, which equals 2y*y', the same thing you get with implicit differentiation.
no because y is a function of x so you have to multiply by the derivative of y with respect to x
A circle can be expressed as 2 functions: y = sqrt(r^2-x^2) (top half) and
y = - sqrt(r^2-x^2)(bottom half). when I do the derivitave of the top half I get
-x/y, just like with implicit differentiation. But when I do the bottom half, I get positive x/y. What am I doing wrong?
You are doing nothing wrong: in the bottemleft quadrant x/y is negative. In the equation x (negative) is divided by y (positive because sqrt( r^2-x^2) stays positive). Example at (-3, -4) the equation gives (-3)/(sqrt(25-(-3)^2)= (-3)/(4). This is true since at the bottomleft quadrant the slope is negative.
very good broseph!
Very helpful
as helpful as your profile pic is when i feel horny to jerk off.
amazing thank you
Thanks amazing video!
why is the gradient negative though?
Thank's a lot.
Aren't there videos for partial differentiation?
Thanks eddie
Edie woo makes me me go edie wow
Thank you teacher
at 10:19 why did you change (d/dx)y to dy/dx, because aren't you essentially going from the derivative of y wrt x to the rate of change of y to x which are two different things?
Just one question: how would you go about graphing this implicit gradient?
Thank you, you are amazing
5:27 Woo hoo...
THANK YOU SIR!
Why at 11:03 Prof. Eddie Woo mentions dy/dx as gradient rather than derivative ( He seems to self correct).?
In this context,"gradient" means the actual numerical value at some specific point on the curve.
The derivative is the formula to calculate the gradient.
He already has the derivative. It is -x / y. Now he plugs in the values and gets the actual gradient at the point in question.
Remember: "gradient" is just a fancy word for "slope of the tangent". But this is a numerical value. The derivative is a formula to calculate this slope at any point on the curve. The derivative describes in which way this slope changes when walking along the curve.
great
I would have liked one more example with a y impossible to extract...
He made 2 errors. First the sq. root is +/- and not just + making the first method more tedious. Second one should show all the steps of the chain rule making the second method look easier and the preferable method since one CANNOT always make Y explicit. So when trying to find dY^2/dX by chain rule he should show all the steps. So initially we should write d Y^2/dY and not dY^2/dX. Then one can easily see that dY^2/dY = 2Y and dY^2 = 2Y*dY. Then dividing both sides with dX we get dY^2/dX = 2Y*dY/dX.
sir i am not a mathematician and i just want to understand dx/x integral could you make a video about it? thank you... im just so curious
he has made videos on it
He sounds pained at the start of the vid
👋👋
For explanation of why its called implicit differentiation and not just normal differentiation go here : th-cam.com/video/wcn9r64X5bQ/w-d-xo.html
does anyone else answer his questions? xD like in the video when he asked if we were happy with how he got to the solution, i was like "wait *pauses* *scans the board* okay yea i got it". xD
There's no second part,right?