 Remember, we can express the derivative using operator notation. So the derivative of f with respect to x can be written df dx, so we write the operation of differentiating with respect to x as d over dx. We can then apply this to f to get the derivative. d over dx applied to f does give us df dx. Now in a similar way, we can define the partial differential operators, partial with respect to x, partial with respect to y, with respect to z, and so on. And we define the vector differential operator, this inverted delta, pronounced del, to b, where i, j, and so on are the elementary vectors. Now this symbol, which we usually read as del, is officially called nabla because of its resemblance to a harp. If f is a multivariable function, we say that nabla f or del f is the gradient of f. For example, let's f of x, y be this function of three variables, and let's find and interpret the gradient of f. So del f, well that's our vector differential operator applied to f, so we'll preface it. Now you might be a little bit worried about what's in this dot, dot, dot, but let's go ahead and write it down anyway and then worry about it when we get to it. One of the reasons operator notation is convenient is that we can use these operators and treat them almost as if they were factors, and so f is acted on by every one of these operators. And so we want to find the partial of f with respect to x, y, z, and more variables. So we can find the partials with respect to x, y, and z. Hidden in this dot, dot, dot are things like the partial of f with respect to w. Well, let's find out what that is. And it should be clear why the dot, dot, dot doesn't matter. These represent the partial derivatives with respect to variables that f doesn't contain, and so all of those later derivatives are going to be zero and will drop out. The only things that do stay are the partial with respect to x, y, and z, and because they're multiplied by the elementary vectors i, j, and k, we can rewrite them as a single vector and substitute in our partial derivatives. And it's worth noting that the vector that we're actually finding is the vector whose components are the partial derivatives with respect to x, y, and z, and you might remember this is the direction of greatest increase of our function from a point. Now let's consider the curve f of x, y equals c. The tangent line to this curve at some point will have a slope we can find using implicit differentiation, and here we'll need to use that multi-variable chain rule. Now remember that the line with slope a divided by b runs in the same direction as the vector b, a. So we can represent our slope as a vector, and so the tangent line points in the same direction as the vector. Meanwhile, if we find the gradient of f, so remember that's our vector differential operator applied to f, and since f is a function of just the variables x and y, only the partial derivatives with respect to x and y are relevant. Applying our operator gives us, which we can write in vector form. Now we have two vectors, and as a useful rule and strategy in higher mathematics, if you have two vectors, find their dot product. Well sure, why not, we have nothing better to do, so let's find the dot product of the two vectors, which is zero, and remember if the dot product of two vectors is zero, then the vectors are orthogonal. And this means that the gradient of f is orthogonal to the tangent line. We say that the gradient of f is a vector normal to the curve, and this gives us another useful and relatively easy thing to find, find the vector normal to the curve at a point. Now again, it's useful to remember that we should always try out new methods on problems where you know what the answer is supposed to be, and so keep in mind that for this problem we can do this in two ways. We can find the tangent line, then find the perpendicular line, or we can use the gradient, and we'll do it both ways. And so using implicit differentiation, we find the slope of the line tangent to the graph, which will be negative one. Now we want the vector normal to the line, and so we want the perpendicular line. So remember the slopes of perpendicular lines are negative reciprocals, so if the slope of the tangent line is negative one, the slope of the perpendicular line will be negative the reciprocal. And so the perpendicular line will have a slope of one, and let's convert that into a vector. Remember that a line with a slope of p over q has the same direction as the vector qp, and so this slope one is going to correspond to the vector one one. Now we can also use the gradient. So notice that our curve is the f of xy equals negative one level curve of our function x cubed plus y cubed minus 2xy. And so we can find the normal vector by finding the partial derivatives, which will be, and we get the same answer as before for the vector normal to the curve.