Compute the directional derivative of f at the given point in the direction of the indicated vec

Compute the directional derivative of f at the given point...

Key Concepts

Unit Vector

A unit vector is a vector with a magnitude of one. When calculating directional derivatives, it is necessary to normalize the given direction vector to ensure that the derivative reflects the rate of change per unit distance. This normalization ensures that the directional derivative is independent of the magnitude of the chosen direction vector.

Directional Derivative

The directional derivative measures the rate at which a function changes at a point in a specified direction. It is computed as the dot product of the function’s gradient and a unit vector in the desired direction. This concept generalizes the idea of a derivative to directions other than the coordinate axes.

Implicit Function Theorem

The Implicit Function Theorem provides the conditions under which an equation involving several variables implicitly defines one variable as a differentiable function of the others. This theorem assures that if certain non-degeneracy conditions are met (for example, a non-zero derivative with respect to the variable being solved for), a local explicit form exists, permitting the use of differentiation techniques on that function.

Implicit Differentiation

Implicit differentiation is a technique used when a function is defined implicitly by an equation rather than explicitly. It allows us to differentiate both sides of the equation with respect to a variable, treating other variables as functions and applying the chain rule accordingly. This method is essential when computing partial derivatives for functions defined by relations among several variables.

Gradient

The gradient is a vector consisting of all the first-order partial derivatives of a function. It points in the direction of the greatest rate of increase of the function. In the context of directional derivatives, the gradient provides a compact way to encapsulate how the function changes with respect to each independent variable.

Transcript

00:01 All right, in this video, we are going to find the directional derivative of c, which is a function of x and y, in the direction of 3 -1 -1 at the point 1 comma 1.

00:11 So the first thing we'll do is we're going to compute the unit vector, because what we essentially need to do is you need to find the gradient in the unit vector and dot them at the point 1 -1.

00:25 So our unit vector is going to be vector u divided by its magnitude, and that in the end gives you 3 over root 10, negative 1 over root 10.

00:47 Okay, and now what we need to do is we're going to use implicit differentiation to find the partial derivative of z with respect to x and the partial derivative of z with respect to y.

01:04 So we're going to have to do those each separately.

01:06 So let's do, right, so remember our gradient, right, our gradient here, just as it reminds.

01:15 Is f partial x, f partial y, and of course z is f, so we're finding z partial x, z partial y.

01:24 So let's do fx first.

01:29 So i'm going to have to take the partial derivative with respect to x.

01:34 That means i'm holding y constant, and then i'm differentiating z implicitly.

01:42 So the derivative of 2x squared y would just be 2x squared, 2x squared.

01:49 Sorry, it would be 4xy.

01:53 Okay, the derivative of negative y z cubed would be negative 3y z squared times partial z, partial x.

02:06 I want to use the product rule here.

02:08 The derivative of x is 1 times z squared, plus x times 2z, partial z, partial x.

02:21 And the derivative of 2x squared with respect to x is, i'm sorry, 4x.

02:30 Okay, that gives you a 0 at the end.

02:32 So now we just collect the terms that do not have a derivative involved.

02:37 That's this, this, and this.

02:39 I'm going to throw those on the other side of the sql sign, and that gives me a 4x minus a 4xy minus a z squared, divided by and everything that has a derivative attached will end up being divided.

03:06 So in one fell swoop here, that ends up being 2xz minus 3yz squared.

03:16 And there is partial z, partial x.

03:25 All right, so we're going to need to come back to that and evaluate it later.

03:31 So i'll just keep that in mind, but now recall i need to do this again to find my partial derivative with respect to y.

03:45 So same process.

03:46 This time we hold x constant.

03:52 Okay.

03:52 And so this derivative here, holding x constant, becomes 2x squared...