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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.