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$$\begin{aligned}&\text { Suppose } f(x, y)=3 x^{2} y^{2} \text { determine (a) } \lim _{h \rightarrow 0} \frac{f(x+h, y)-f(x, y)}{h}\\&\text { (b) } \lim _{k \rightarrow 0} \frac{f(x, y+k)-f(x, y)}{k}\end{aligned}$$

(a) $6 x y^{2}$(b) $6 x^{2} y$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 3

Limits and Continuity

Derivatives

Campbell University

Baylor University

Boston College

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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Hello in the question the function it is given that f, x, comma y, that is, equals to x square plus 2 y square. Okay and the part is limit, h, tends to 0 limit, tends to 0 f x, minus f, f x y upon h, and we know that it is equal to delta, f x, comma y delta x, partial derivative with respect to x. So this is the function. F, x y, so this is the function it is given in the question, and now we will partially differentiate it with respect to x. Holding by is the constant, so it will be cos to 2 x and the part b it is asked limit x, tends to 0 f x y plus k minus f of x, comma y, upon h, that is delta, f, x, comma y upon delta y. It is the partial differentiation of x, with respect to y holding x, as a constant set will be equals to 4 y. Sorry here, k is here since just given k so ka tends to 0 upon k. Okay, so it is defined by delta, f by delta y equals to so we'll differentiate, partial differentiation of this with respect to y holding x as a constant, so it is equals to 4 y. These are the answers. I hope you understood. Thank you.

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