Let x3 y3 5z3 5xyz 210 0 use partial derivatives to...

Transcript

00:01 In this video we are given that negative 2x cubed minus y cubed plus 5z cubed minus 5xyz minus 894 equals 0.

00:15 We want to find the partial of z with respect to x and the partial of z with respect to y at the point 1, 4, 6.

00:24 Okay, so we're going to let f of x y z equal negative 2x cubed minus y cubed plus 5z cubed minus 5xyz minus 894.

00:38 The partial of f with respect to x is negative 6x squared minus 5yz.

00:45 The partial of f with respect to y is negative 3y squared minus 5xz.

00:50 The partial of f with respect to z is 15z squared minus 5xy.

00:58 So the partial of z with respect to x is negative of the partial of f with respect to x divided by the partial of f with respect to z, which is equal to negative, that's on the outside, and then we have negative 6x squared minus 5yz divided by 15z squared minus 5xy.

01:24 And the partial of z with respect to y is negative of the partial of f with respect to y divided by the partial of f with respect to z, which is negative of, and then we have negative 3y squared minus 5xz divided by 15z squared minus 5xy.

01:48 So now that we have the partial of z with respect to x and the partial of z with respect to y, we could evaluate both of those at the point 1, 4, 6.

02:00 So if i want the partial of z with respect to x at the point 1, 4, 6, i'm going to come up here and replace x with 1, y with 4, and z with 6.

02:14 And you might notice i have a sum in the numerator.

02:19 So what i did was i distributed the negative to each term within the numerator.

02:24 And a negative times a negative is a positive, right? so the partial of z with respect to x evaluated at the point 1, 4, 6 is 6 times 1 squared, right? replace x with 1, plus 5 times 4 times 6...

Question

Let $x^3 + y^3 - 5z^3 - 5xyz + 210 = 0$. Use partial derivatives to calculate $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ at the point $(-5, 5, -2)$.

Question

Let $x^3 + y^3 - 5z^3 - 5xyz + 210 = 0$. Use partial derivatives to calculate $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ at the point $(-5, 5, -2)$.

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