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Find the partial derivatives with respect to (a) $x,$ (b) $y$ and (c) $z$.$$f(x, y, z)=3 e^{2 x^{2} y^{3} z^{4}}$$

(a) $12 x y^{3} z^{4} e^{2 x^{2} y^{3} z^{4}}$(b) $18 x^{2} y^{2} z^{4} e^{2 x^{2} y^{3} z^{4}}$(c) $24 x^{2} y^{3} z^{3} e^{2 x^{2} y^{3} z^{4}}$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 2

Partial Derivatives

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Lectures

12:15

In calculus, partial derivatives are derivatives of a function with respect to one or more of its arguments, where the other arguments are treated as constants. Partial derivatives contrast with total derivatives, which are derivatives of the total function with respect to all of its arguments.

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Find the partial derivativ…

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Find the first partial der…

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Find the indicated partial…

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So for us to find the partial derivatives of this year, remember, what we're going to do is assume all other variables are going to be a constant. And then we just take the derivative just like we would in the single variable case. So if I were to do del by Dell X here to try to find the partial with respect to X, we would write this as DLF by Dell X is equal to f sub X. And then, well, we would need to use chain rule here, so it would be three e because the derivative of three times e to the something will eat to something is always eating something times the derivative of the power. So just be two x squared y cube Z to the fourth. And then we have Dell by Dell x of two X squared y Cube Z to the fourth. Now, remember, we're assuming that Y and Z are constants, so we can factor those out. I'll just pull those to the front. So three y q z to the fourth e two x squared like you see to the fourth. And then we have just Dell by Dell acts of two X squared. Oh, and actually, I guess I can pull this to out as well because that's a constant. So that would just become six out front. And now we could take the derivative of X squared just using powerful. So that would be two X. And then when we multiply everything together, we get 12 x. Why cubed Z to the fourth e to the two expert y Cube See to the fourth. So that is our partial respect. Two X right? Next, we could go ahead and take the partial respect to why? Which is going to be the same process? Just the same. We assume that X and Z are constants. We have DLF by Dell. Why or we can write is f sub y. So again we'll start the same way three times e to the two x squared y Cube, See to the fourth and then Del by Dell Nazi and ahead of myself, why of two X squared y Cube Z to the fourth. And I remember we're now going to assume that X and Zeer constant so we can pull those out and I'll pull the two out as well So it be six um X squared Z to fourth e to the two x squared y cube to the fourth. And then we just have Dell by del y of White Cube. And then to take the derivative of that we will use powerful. So that would be three y squared that we could multiply everything together. And this is going to give us 18 x squared y squared C to the fourth e to the two x squared y que si to the fourth. And that is our partial respect to why. And it will be the same thing for find the partial respect to Z. But now we assume X and why are going to be our constants. So Dell, by Del Z on each sides that give us DLF by del Z or F sub Z is equal to so would be three e to the two x squared y cube T to the fourth and then del by del Z of the insights of two X squared y cubes Fourth. But now this acts of this wire assuming or constant, so we can pull those to the front. So it would be six x squared y cube e 22 x squared y que si to the fourth del by del C of C to the fourth. And then we can take the derivative that using powerful. So that would be four z cute that we could multiply all that together. So we have 24 x squared y cute Z cube e to the two x squared y cube Z to the fore. And then this here is going to be our partial derivative with respect to Z. So again, remember, we take these partials just like we would any other derivative, but we just assume all other variables are constant.

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