Each factor of this journal is packed choked with step by step initiatives that flip the normal into amazing. Australian Handyman supplies nice how to increase either your outdoors and indoor region, it informs you approximately tendencies and tasks that might decorate your house, prevent funds and while supply your house that particular suppose of being your house.

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Example text

Quiz Problem 17 1. Show that the function f (x, y) = x sin y x 2 + y2 does not have a limit as (x, y) → (0, 0). 2. Is the function f (x, y) = x+y x+y 1 (x, y) = (0, 0) (x, y) = (0, 0) continuous at (0, 0)? 3. Find the domain of the function f (x, y) = ln 1 . 1 Partial Derivatives What shall we mean by the derivative of f (x, y) at a point (x0 , y0 )? Just as in one variable calculus, the answer is the slope of a tangent line. The problem with this is that there are multiple tangent lines one can draw to the graph of z = f (x, y) at any given point.

X xy Problem 19 Compute ∂f (x, ∂x y) and ∂f (x, ∂y y) for the following functions. 1. x 2 y 3 2. xy Problem 20 For the function f (x, y) = −x + x y 2 − y 2 find all places where both ∂f and ∂∂ yf are zero. 1 COMPOSITION WITH PARAMETERIZED CURVES Suppose we have a parameterized curve φ(t) = (x(t), y(t)) in the plane. That is, for a given value of t we are given the numbers x(t) and y(t), which we visualize as a point in the plane. We can also take these two numbers and plug them in to CHAPTER 3 Derivatives 27 a function f (x, y).

There might be some way to approach (a, b) that you haven’t tried that gives a different number. This is the key to the definition of limit. We say the function has a limit only when the values of f (x, y) approach the same number no matter how (x, y) approaches (a, b). We illustrate this in the next two examples. EXAMPLE 2-2 Suppose f (x, y) is given by f (x, y) = x2 xy + y2 As we let (x, y) approach (0, 0) along the x-axis (where y = 0) we have f (x, y) = x2 0 xy = 2 =0 2 +y x Similarly, as we let (x, y) approach (0, 0) along the y-axis (where x = 0) we have f (x, y) = x2 xy 0 = 2 =0 2 +y y But if we let (x, y) approach (0, 0) along the line y = x we have f (x, y) = So once again we find lim (x,y)→(0,0) xy x2 1 = = 2 2 2 x +y 2x 2 f (x, y) does not exist.

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