3.4: Partial Derivatives Definition Mat 22-Lecture 9 For a single-variable function z = f(x), te derivative is f (x) = lim 0 f(x+) f(x). For a function z = f(x, y) of two variables, to define te derivatives, we old one variable fixed: f x (x, y) = z x = f x = lim 0 f y (x, y) = z y = f y = lim 0 f(x +, y) f(x, y), f(x, y + ) f(x, y). Tey are called te partial derivatives. We can similarly define partial derivatives for a function of tree variables w = f(x, y, z). Interpretation (Draw te picture.) Instantaneous rates of cange wit respect to x or y Slope of te line tangent to te x-curve or y-curve. (We fix y = y 0, te curve z = f(x, y 0 ) is called an x-curve.) Example: consider f(x, y) = { xy x 2 +y 2 (x, y) (0, 0) 0 x = y = 0 Te function is not continuous at (0, 0) but we can compute te partial derivatives at (0, 0) (tis means tat te function is continuous along x and y directions but not continuous along an arbitrary pat.). Find f x (0, 0) and f y (0, 0). f(, 0) f(0, 0) 0 0 f x (0, 0) = lim = lim = 0. 0 0 Similarly, f y (0, 0) = 0. Comment: In te one-variable function case, te existence of derivative is equivalent to differentiability, wic implies continuity. By tis example,
for multi-variable case, te existence of partial derivatives doesn t imply te continuity of te function itself. It only implies tat te function is continuous along x or y direction. Later, we ll see tat differentiability implies continuity. Wen we compute te partial derivatives, te oter variable is regarded as a parameter (constant). Terefore, te product rule, cain rule etc for one-variable functions still old for partial derivatives. Example: Consider w = f(x, y, z) = xyz ln(xyz). Compute f x and f z. f x = (xyz) x ln(xyz)+(xyz)(ln(xyz)) x = yz ln(xyz)+xyz yz = yz(ln(xyz)+). xyz Similarly, you can compute f y and f z. Example: Suppose z = f(x sin(y/x)) were f is a smoot one-variable function. Compute z x. By te cain rule for one-variable, we ave z x = f (x sin(y/x))(x sin(y/x)) x = f (x sin(y/x))[sin(y/x)+x cos(y/x)( y/x 2 )] =.... Example: Consider te function { xy (x, y) (0, 0) f(x, y) = x 2 +y 2 0 x = y = 0 again. Compute f x (a, b) and f y (a, b) for (a, b) (0, 0) and conclude tat f x, f y are not continuous at (0, 0). Hig order derivatives Clairaut s teorem: f xx = x (f x) = 2 f x 2 f xy = y (f x) = 2 f y x f yx = x (f y) = 2 f x y 2
Teorem. If bot f xy and f yx are defined in a ball containing (a, b) and tey are continuous at (a, b), ten f xy (a, b) = f yx (a, b). If tey are not continuous, it s possible tat tey are not equal (see Problem 74). Proof. f xy (a, b) = lim k 0 k (f x(a, b + k) f x (a, b)) = lim k 0 k lim 0 ([f(a +, b + k) f(a, b + k)] [f(a +, b) f(a, b)]) Now, using tat f y is defined in te neigborood, applying te mean value teorem to g(y) = f(a +, y) f(a, y), we ave lim lim k 0 0 [f y(a +, b + ξ(k)) f y (a, b + ξ(k))] Applying te mean value teorem again, lim lim f yx(a + η(), b + ξ(k)) k 0 0 using te fact tat f yx is continuous, we are done. Example: Given P = 2x + cos x sin y and Q = sin x cos y + e y, can you find f suc tat f x = P and f y = Q? If it s impossible, explain wy; if it is possible, find suc one f. Suppose suc one f exists. Ten, f xy = P y = cos x cos y, f yx = Q x = cos x cos y. Bot of tem are continuous. Hence, if te f exists, we must ave P y = Q x. Tis is true for tis example. Hence, f sould exist (tere is anoter ting about te domain, but we ll come back to tis very later). Ten, to recover f, f = P dx = x 2 + sin x sin y + C(y). 3
Ten, f y = Q tells us tat sin x cos y + C (y) = sin x cos y + e y. ence, C (y) = e y or C(y) = e y + C. Hence, f(x, y) = x 2 + sin x sin y + e y + C. Anoter way is to use te differentials to find f. Example: Suppose te ig order partial derivatives of f(x, y, z) are all continuous. Sow tat f xxz = f zxx. Comment: If te derivatives are continuous, te order of taking te derivatives is not important. Proof. First of all, let g = f x. Since te derivatives are continuous, by te teorem, g xz = g zx or f xxz = f xzx. Now, as functions, f xz = f zx by te teorem again. Hence, f xzx = f zxx. 3.6: Increments and linear approximation In te previous lecture, we see tat te existence of partial derivatives f x, f y doesn t imply tat te function is continuous. However, if f x and f y are continuous at point (a, b), ten f is continuous at (a, b) and furtermore, we ave te estimate for f(a + x, b + y) f(a, b): Consider te increment f = f(a + x, b + y) f(a, b). We ave by te mean value teorem: f = [f(a + x, b + y) f(a + x, b)] + [f(a + x, b) f(a, b)] = f y (a + x, b + ξ( y)) y + f x (a + η( x), b) x Since f x, f y are continuous at (a, b), R = f y (a+ x, b+ξ( y)) f y (a, b) 0 and R 2 = f x (a + η( x), b) f x (a, b) 0. Hence, f = f x (a, b) x + f y (a, b) y + R x + R 2 y. From tis expression, f is continuous at (a, b) and we ave te linear approximation of f(x, y) near (a, b): f(x, y) f(a, b) + f x (a, b)(x a) + f y (a, b)(y b). Tis is good if x a and y b are small. Example: Compute e 0. 7 approximately. 4
Since 7 = 4 + /6. We can terefore define a function f(x, y) = 4e x + y. We are supposed to compute f(0., /6). f(0, 0) = 4 f x (x, y) = 4e x + y f x (0, 0) = 4 f y (x, y) = 2e x + y f y (0, 0) = 2. Hence, f(x, y) 4 + 4 0. + 2 6 = 4.525 5