Hello, wonderful mathematics people. This is Anna Cox from Kellogg Community College. Implicit differentiation. First we're going to differentiate both sides of the equation with respect to X. Then we're going to collect all the DYDX terms on one side, and then we're going to solve for DYDX. So the first thing we're going to do is we're going to take X, the first example, X ^3 minus XY plus y ^3 equaling 1. So we're going to differentiate both sides of the equation with respect to X. So X ^3, the derivative is just three X squared, the X * y two different variables, they're both changing. So the derivative of X is one times the Y plus the derivative of plain Y is going to be DYDX or Y prime times the first plus y ^3. Well, we're going to take the derivative of y ^3 in terms of X. So we're going to get three y ^2 d YDX or Y prime equaling the derivative of a constant which is always 0. So in reality, we're just taking the derivative of every single variable in terms of X. Sometimes students like to think of taking the derivative of DX DX, which is really one. So if we started back here with the original X ^3, 3 X squared derivative of X in terms of X, then minus the XY, the derivative of X in terms of X is just one or DX DX times the Y plus the derivative of Y in terms of XDYDX times the 1st or X plus y ^3 would be 3 Y squared, the dy in terms of the X equaling 0. So now we're going to distribute and we're going to combine and we want to get all the dy DX's on one side by themselves. So the DX over DX is just really one. So the dy DX negative X + 3 Y squared, taking the other terms to the other side, so y -, 3 X squared dividing by what was multiplied by the dy DX and we get dy DX equal negative Y minus three X ^2 / -X + 3 Y squared. We don't Care now that we have both X's and YS in the equation because eventually when we evaluated a location, we're going to be able to find both X&Y as a point. We want to talk about a normal, the fact that a normal line is perpendicular to the tangent line. So in point slope form, you're going to have Y equaling B plus the perpendicular DYDX at the point AB of the quantity at times the quantity X -, a. So that's really just point slope form. The power rule for rational powers. If P / Q is a rational number, then X to the P / Q is differentiable at every interior point of the domain of X to the P / Q - 1, and the derivative of X to the P / Q equal P / Q * X to the P / Q - 1. The proof. Let's start with letting PQ be integers and Q has to be positive. So if we started with the original Y equal X to the P / Q, and we took each side to the Q power, we'd then have yq equaling X to the P. We're going to take the derivative of each of those. So we get qy to the Q - 1, dy DX equaling PX to the P - 1, and we could think of that as DX DX, which is just one. So dy DX is going to be PX to the P - 1 divided by qy to the Q - 1. Now remember back in the beginning, we let Y equal X to the P / Q. So if we substitute, we get PX to the P - 1 / Q instead of Y. We're going to put in XP to the Q all to the Q - 1. If we distribute that out powers to powers, we multiply. So P / Q * Q is P P / Q * -1 is negative PQ. If the bases are the same and we're dividing, we subtract the exponents. So we get P / Q * X to the P - 1 -, P + P / Q. The PS are going to cancel and we get the derivative of Y in terms of X equaling P / Q of X to the P / Q -, 1. Doing some examples, we're going to be given an equation and a point. So the first thing we're going to do is find the derivative. We need to use implicit differentiation because we have both X's and Y's now. So the derivative of X ^2 is 2X the derivative of negative root 3 XY. We're going to take the derivative of one of the terms, the Y. So Y prime times the rest minus the derivative of the X which was just one times the rest. So minus root 3 / y + 4 YY prime equaling 0. Collect our Y primes all on one side, everything without AY prime on the other. Factor out the Y prime and divide. Now we're doing it at a specific point. We're doing it at root 3, 2. So if we put root 3 every time we see an XN and we put 2 every time we see AYN and we compute in this case we get a derivative of 0. What kind of line gives us a derivative OF0A horizontal line? So our horizontal line or our tangent is going to be Y equal the Y value of two. Our normal, in this case perpendicular to a horizontal line is vertical. So it's going to be X equal root 3. Thank you and have a wonderful day.