 So we found that the plane tangent to some surface z equals f of x, y at some point will have a normal vector whose components are the partial of f with respect to x with f with respect to y, negative 1. But what happens if z is defined implicitly? And here it's important to remember a problem exists whether or not you can solve it. So let's see how we might solve it. So remember that our ability to find the tangent plane relied on our ability to find two tangent lines which meant we needed to find the derivatives. So remember we obtained dz dx by holding y constant and differentiating normally. We can do the same thing here if we use implicit differentiation. So if we assume that y is a constant, then we can differentiate with respect to x, then solve for dz dx. And since we know the values of x and z, we can substitute those in to get the value of dz dx. So again the derivative is the slope of the tangent line and so we can think about this as rise over run. So we have a run that's along the x-axis of negative 3, so our first vector component will be negative 3. We're assuming y is constant, so our second component is going to be 0, our rise is 5, and we can take that as our third component. And so one vector in the tangent plane is negative 3, 0, 5. And similarly if we assume x is a constant, we can differentiate implicitly with respect to y and find dz dy. And again derivative is slope, slope is rise over run, and so this derivative 8 divided by negative 3 corresponds to a vector pointing in the direction of the tangent line. x is constant, so the first component of the vector is 0, our run is minus 3 which runs along the y-axis, so our second component is negative 3, and the rise is 8 along the z-axis. And so that will be our third component, and so our second vector that is in the tangent plane, 0, negative 3, 8. And now that we have two vectors in the tangent plane, we can find the normal vector by taking the cross product. And once we have the normal vector at a point in the plane, we can write the equation for the plane. Let's think about this in a different way. Our surface z squared equals x squared minus 4y squared, well I can rearrange the equation. And if I introduce a new function of 3 variables, then we get the surface corresponding to the equation 0 equals f of x, y, z. And what this means is that the surface z squared equals x squared minus 4y squared is a level curve, well actually now it's a surface, a level surface of our function of 3 variables. And specifically it's a level surface where our function of 3 variables is equal to 0. So suppose a function of 3 variables defines a level surface. Now if we want to find the tangent plane, we need to find the derivative of z with respect to x and the derivative of z with respect to y. So originally to find dz dx, we assumed y constant and used implicit differentiation. And so we'll do that here as well. So the equation is f of x, y, z equals c. And there's three ways that this function might change when we change x. First it might change because of the first variable, so we need the partial of f with respect to x times the chain rule always applies the derivative of x with respect to x. But wait, there's more. This function could also be changing with respect to y. So we want the partial of f with respect to y times, chain rule always applies, dy dx. And again some of the variation in the function might come from that third component. So that's our partial with respect to z times, don't forget the chain rule, dz dx. Meanwhile over on the right hand side c is a constant and it never changes, so our right hand side is going to be 0. Now remember that when we found dz dx we held y constant. So if we're holding y constant, dy dx has to be 0. And dx dx, well one of the advantages of differential notation is that some things are exactly what we expect them to be. dx dx is equal to 1 and so this means we get an equation that we can solve for dz dx. Similarly if we hold x constant we can use implicit differentiation to solve for dz dy, so our first step. Since x is constant, dx dy is equal to 0, dy dy is equal to 1 and so we find. Since we have the derivatives we can translate these into vectors in the tangent plane and so we can take the cross product. Notice that every component of our vector has this factor of the partial of f with respect to z, so we'll remove that common factor. And remember that any scalar multiple of a vector points in the same direction. So we can actually just drop that common factor and get a vector normal to the tangent plane. And again somebody says that here's a brand new method for doing things, always check a new method in a problem where you already know the answer. So we've already solved this problem. First we'll rewrite the equation defining our surface so we can describe it as a level surface. And that's an important step and the only challenging part here is getting all of our variables on one side of the equals. Okay maybe it's not that challenging but that does give us the function of three variables. And so our surface is the f of x, y, z equals zero level surface of our function of three variables. The plane tangent at the point will have a normal vector whose components are the partial derivatives. So let's find those and we get the equation and that's different. Well that's not panic just yet. When we solved this problem earlier we obtained this as our equation. But this was because we used the normal vector negative 15, 24, negative 9. And we got this equation because we used the normal vector 10, negative 16, 6. Now the thing to notice here is that the vector that we used, well that's the scalar multiple three times negative 5, 8, negative 3. Our new normal vector is really the scalar multiple 2 times 5, negative 8, 3. Or factoring out another negative 1, negative 2 times negative 5, 8, negative 3. And so the two normal vectors are scalar multiples of each other. And that means while our equations are different they are equivalent equations.