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In Exercise $2,$ determine (a) $\lim _{h \rightarrow 0} \frac{f(x+h, y)-f(x, y)}{h}$,(b) $\lim _{k \rightarrow 0} \frac{f(x, y+k)-f(x, y)}{k},$(c) $\lim _{h \rightarrow 0} \frac{f(2+h, y)-f(2, y)}{h},$(d) $\lim _{k \rightarrow 0} \frac{f(x, 3+k)-f(x, 3)}{k}.$

(a) $2 x-9 x^{2} y$(b) $-2 y-3 x^{3}$(c) $4-36 y$(d) $-6-3 x^{3}$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 1

Functions of Several Variables

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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01:17

For $f(x, y)=2 x y^{2},$ f…

01:58

For $f(x, y)=x^{2}+2 y^{2}…

05:56

Let $f(x, y)=4 x^{2}-2 y^{…

00:59

Given $f(x, y)=x^{2}-4 y$ …

you know the function F x y it is given that two x Y squared okay. And for the party limit agents to zero. If explains what have explosives come away? Monies of f x y upon edge. So we know that it is a cost to and they'll have won their legs. It is a partial differentiation of X f with respect to X, so it will be cause to We will buy as a constraint. So it will be two y squared for the party it has asked LTD caters to zero f of x y plus key upon into sorry negative of have X comma y upon a key that isn't the case tends to zero. So it is different by the left x comma y upon Dell way. It is a positive differences in of health with respect to y holding X as a constant. So it will be a cost to full X y This is the answer. I hope you are national. Thank you

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