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Let $f(x, y)=3 x^{2} y^{3} .$ Show that $f(x+\Delta x, y+\Delta y)-f(x, y)=$ $f_{x}(x, y) \Delta x+f_{y}(x, y) \Delta y+\epsilon_{1} \Delta x+\epsilon_{2} \Delta y,$ where $\epsilon_{1}$ and $\in_{2}$ approach zero as $\Delta x$ and $\Delta y$ both approach zero.

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 function values.<…

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Find the indicated quantit…

01:57

Let $f(x, y)=\left(x^{2}+y…

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Find and simplify the func…

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Verify that $f_{x y}=f_{y …

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02:02

Let $f(x, y)=x y+3 .$ Find…

for this problem, we are asked to find the function values for F. Of X Y equals three X. Y plus Y squared. For part A. We want F of X plus delta X. Y. So all we need to do is plug in X plus delta X. Where we see X in the original statement of the function so F of X plus delta X Y is going to be three times X plus delta X times why plus Y squared? We could expand this out further, but it's not really going to do anything useful for us for part B. We want F F X, Y plus delta Y minus F F X. Y over delta Y. So now we just need to plug in Y plus delta Y. Where we had Y in the original equation. So we have three X times Y plus delta Y plus Why? Plus delta Y squared minus three X, Y minus Y squared. Right. All divided by delta Y. Uh huh. So we can do some meaningful simplification here. First of all, when we expand out the Y plus delta Y squared, we know that we'll have a Y squared And then we'll also have a two Y Delta Y. And then we'll have a delta Y squared. So when we have that that plus Y squared from expanding of those brackets that will add with that minus Y squared up to zero. So we lose that term. We have a plus three X. Y. And a minus three X. Y. So those will add up to zero. So we'll be left with on top three X. Delta Y. Uh Plus to my delta Y plus delta Y squared all over delta Y. Which will then equal three X. Three X. Plus two Y. Plus delta Y. Yes.

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