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.
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.
For explanation of why its called implicit differentiation and not just normal differentiation go here : ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-wcn9r64X5bQ.html
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.
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.
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?
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 !
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.
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
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.
