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Find the partial derivatives with respect to (a) $x,$ (b) $y$ and (c) $z$.$$f(x, y, z)=\ln \left(x^{2}+y^{2}+z^{2}\right)$$

(a) $\frac{2 x}{x^{2}+y^{2}+z^{2}}$(b) $\frac{2 y}{x^{2}+y^{2}+z^{2}}$(c) $\frac{2 z}{x^{2}+y^{2}+z^{2}}$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 2

Partial Derivatives

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

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 indicated partial…

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Find the first partial der…

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So if we want to find the partial derivative of this, remember, what we're going to do is just take the derivative like we normally would. But then all other variables, we assume our constant so we would treat them just like any other constant in the single variable case for directives. So, first, would you tell by Dell X So that would give us Del F by Dell X, which will often right is F sub X and then over here to take the derivative of natural log of something we first would do one over the thing on the inside, and then we apply chain rule. So we're going to do del by Dell x of what was on the inside. So expert plus Weiss. We're plus disease word. And I remember this why and this Z here we assume there constant. So if I square constant, they're still constant. And if I take the derivative of Constant, that's just going to be zero in each case. And then to take the euro X where we just use power rules, that will be two X so that would give us two x over X squared plus y squared plus z sward. Um, now to do the partial would respect why will go through the exact same process. But now we're going to assume X and Z are constants. So we do del by almost forgot to write. Uh, this was our partial with respect to X here. Okay, now we dio del by del y of this So that would give us LF, Beidle why is f sub y so again? Take the derivative of the upside. We're just gonna apply chain will so they'll still be blown over x squared plus y squared plus z squared. And then we do del Beidle Why of what's on the inside due to change? Well and now X and Z, we're assuming our Constance So that is going to be zero for the same reasons before. And this c squared is also going to be zero when we take the derivative. And then why square This is going to be two y So this is going to give us so are partial with respect. Why is to why, over x squared plus y squared plus disease word? Yeah, and so that's our partial respect life. And then lastly to get our partial perspective. See, you might even be able to guess what this is going to be. Because we had two x over X squared plus y squared. Plus these words, same thing here, but just to buy the new mayor. So you might expect this to be a to Z over that, Um so a lot of times when we have this kind of like symmetry in the derivative, stuff like this will pop up when we have partials. Um, well, let's just go ahead and double check to see if that is true. So we applied Del by del Z on each side. So we have d f over d Z or DLF by Del Z is equal to half subsidy. So again, um, chain rule. So it would be one over X squared plus y squared plus C squared. And then del by del Z of X squared plus y squared plus the squared. So now we assume X and Y are going to be constant, so squaring constant so constant. So if I take the derivative, these those air just zero and then x squared becomes to Z, then we go ahead and just multiply all those together. And that is going to give, um, to Z over X squared plus y squared plus z squared. And then this is our partial with respect to Z. So again, remember, when we're going through when taking these partials, just assume every other variables a constant and then treat it just like you would. I'm a single variable. Kevin was problem.

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