21-26 Use the Chain Rule to find the indicated partial derivatives. z=x^2+x y^3, x=u v^2+

21-26 Use the Chain Rule to find the indicated partial...

Key Concepts

Partial Derivative

A partial derivative represents the rate at which a function changes as one particular variable is varied, while all other variables are held constant. It is a fundamental concept in multivariable calculus, where functions depend on several independent variables.

Chain Rule for Multivariable Functions

The chain rule in the context of several variables is a method used to compute the derivative of a composite function. It allows you to differentiate a function with respect to one variable by accounting for the intermediate dependencies and summing the contributions from each path in the composition.

Function Composition

Function composition occurs when one function is applied within another, creating a dependency where the output of one function becomes the input of the next. This concept is key to understanding how changes in one variable propagate through the layers of functions, which is directly addressed by the chain rule.

Evaluation at a Specific Point

Evaluating derivatives at a specific point involves substituting the given values into the derived expressions. This step provides concrete numerical information about the rate and direction of change of the function at that particular point.

Transcript

00:01 So for this problem, we have to find the derivative of z with respect to u, v, and w, given that z is a function of x, y, and that x and y are both functions of u, v, and w.

00:15 So we want to just remember that drawing out this diagram, this tree diagram, is really helpful for us to figure out how to calculate each one of these derivatives.

00:23 So if we're trying to find dzdu, we see that we have dzdx, which is a function of, so z is a function of x and x is a function of u.

00:34 So what we remember is we have to do the derivative of z with respect to x, multiplied by the derivative of x with respect to you.

00:44 And then we add this to the other side, which is dz, d, z, d, y, times d, y to you.

00:54 Something like this.

00:56 And those are our only options.

00:58 So for here, we just calculate these derivatives as we see here.

01:01 So dz, dx is going to just be equal to.

01:04 2 times x plus y to the third, multiplied by d x, du, which is going to be our function here, which is just going to be v squared.

01:17 Next, v squared.

01:20 Next, we have dz, d, z, y.

01:23 So dz, y here, is this going to be 3x, y squared.

01:30 And we leave x as a constant.

01:32 And then we multiply this by d, y, which in this case is just one.

01:38 And so our answer for this derivative is just 2x plus y to the third multiplied by v squared plus 3xy squared.

01:47 And we want to find this for u equal to 2, v equal to 1, and w equal to 0.

01:53 So all we do is we plug in these answers.

01:56 Now just thinking about this quickly, we can also see that x and y are part of our values here.

02:04 So we can find out what x is equal to and what y is equal to at these values.

02:08 So x would be equal to at these values, we would have two for you times one squared plus zero to the third.

02:19 So this means that x is going to be equal to two.

02:23 And then we have that y is going to be equal to two plus one, e to the zero.

02:31 E to the zero is equal to and so we have that this is equal to three.

02:36 And so what we do now is we plug in these values.

02:39 So we have that this is going to be equal to 2 times x, which is going to be 2 plus 3 to the 3.

02:49 V here is equal to 1 plus 3 times 2 times 3 squared.

02:57 And that's all we get.

02:59 So here we would get 4 plus 27 plus...

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