Definition of partial derivative page 128 129 fx

Definition of partial derivative (page 128 -129) f(x, y) = x 2 y + y 3 + 2 y f — = fx = 2 xy x f — = fy = x 2 + 3 y 2 + 2 y ln 2 y z = f(x, y) = cos(xy) + x cos 2 y – 3 z —(x, y) = – y sin(xy) + cos 2 y x fy(x, y) = – x sin xy – 2 x cos y sin y z —(x 0, y 0) = – y 0 sin(x 0 y 0) + cos 2 y 0 x fy(2, /2) = z = f(x, y) = (xy)1/3 y fx = ——— 3(xy)2/3 – 2 sin – 4 cos ( /2) sin ( /2) = 0 x fy = ——— 3(xy)2/3

Compare the functions y = x 2 and y = x 2/3 in R 2, and decide whether or not each function is differentiable at x = 0. Either by looking at graphs or y = x 2 y y = x 2/3 y from the definition of derivative, we see that y = x 2 is differentiable x x at x = 0 but y = x 2/3 is not. Now consider the function z = f(x, y) = (xy)1/3 in R 3. z What does f(x, y) look like in the xz plane where y = 0? z = 0 What does f(x, y) look like in the yz plane where x = 0? z = 0 y x What does f(x, y) look like in the plane where x = y? z = t 2/3 where t represents a variable which gives the value of x = y.

Now consider the function z = f(x, y) = (xy)1/3 in R 3. z What does f(x, y) look like in the xz plane where y = 0? z = 0 What does f(x, y) look like in the yz plane where x = 0? z = 0 y What does f(x, y) look like in the plane where x = y? z = t 2/3 where t represents a variable which gives the value of x = y. x We find that for functions in R 3, it is possible for a “level curve” in one direction to be differentiable at a point while a “level curve” in another direction is not differentiable at the same point. A function in R 2 is differentiable at a point, if and only if there exists a line tangent to the function at the point. A function in R 3 is differentiable at a point, if and only if there exists a plane tangent to the function at the point.

In order that a function f(x, y) be differentiable at a point (x 0 , y 0), the function must be “smooth” in all directions, not just in the x-direction and y-direction. If a function f(x, y) is differentiable at point (x 0 , y 0), then we can use the value of each partial derivative at the point to find the equation of the plane tangent to the function at the point: The equation of the tangent plane can be written as g(x, y) = ax + by + c. The following must be true: g(x 0 , y 0) = ax 0 + by 0 + c = f(x 0 , y 0) , gx(x 0 , y 0) = a = fx(x 0 , y 0) , gy(x 0 , y 0) = b = fy(x 0 , y 0). The equation of the tangent plane is g(x, y) = fx(x 0 , y 0) x + fy(x 0 , y 0) y + f(x 0 , y 0) – fx(x 0 , y 0) x 0 – fy(x 0 , y 0) y 0 which can be written z = f(x 0 , y 0) + fx(x 0 , y 0) (x – x 0) + fy(x 0 , y 0) (y – y 0).

Definition of differentiable and tangent plane in R 3 (page 133) See the derivation on page 132. Example Find the plane tangent to the graph of z = x 2 + y 3 – cos( xy) at the point (8 , – 4 , – 1). f(8 , – 4) = – 1 f — = fx = 2 x + y sin( xy) x f — = fy = 3 y 2 + x sin( xy) y fx(8 , – 4) = 16 fy(8 , – 4) = 48 The equation of the tangent plane is z = – 1 + 16(x – 8) + 48(y + 4) which can be written 16 x + 48 y – z = – 63.

Look again at the definition of differentiable in R 3 on page 133, and observe that we can say f(x, y) is differentiable at (x 0 , y 0) if This is the difference between the exact function value at point (x , y) and the approximated value from the plane tangent to the function at (x 0 , y 0). f(x, y) – f(x 0 , y 0) – fx(x 0 , y 0) fy(x 0 , y 0) (x – x 0) (y – y 0) lim ——————————— = 0 (x , y) (xo , y 0) ||(x , y) – (xo , y 0)|| If x = (x 1 , x 2 , …, xn) is a vector in Rn, and f(x) is a function from Rn to R, then we define f(x) to differentiable at x 0 if f(x) – f(x 0) – Df(x 0) (x – x 0) lim —————— = 0 where x xo || x – x 0 || Df(x 0) is the 1 n matrix of partial derivatives at x 0 , and (x – x 0) is the n 1 matrix consisting of the differences between each variable and its specific value at x 0. (NOTE: A 1 n matrix times an n 1 matrix is the same as the dot product of two vectors.

Suppose x = (x 1 , x 2 , …, xn) is a vector in Rn, and f(x) is a function from Rn to Rm. Then, we can write f(x) = [f 1(x) , f 2(x) , … , fm(x)]. We let Df(x) represent the m n matrix with row i consisting of the following partial derivatives: fi fi fi —— —— … ——. x 1 x 2 xn We call Df(x) the derivative (matrix) of f , and of course Df(x 0) is the derivative (matrix) of f at x 0. Look at the general definition of differentiable on page 134.

Example Find the derivative matrix for w = f(x, y, z) = ( x 2 + xy 4 eyz , ln(xz+y) ). _ _ | | | 2 x + y 4 eyz | 4 xy 3 eyz + xy 4 zeyz xy 5 eyz | | Df(x, y, z) = | | 1 / (xz+y) x / (xz+y) | z / (xz+y) | | |_ _| Find the linear approximation at the point x 0 = (1 , e , 0). x– 1 w = f (x 0) + Df (x 0) (x – x 0) = f (1 , e , 0) + Df (1 , e , 0) y – e z– 0 =

1+ e 4 1 1 + e 4 1 2 + e 4 4 e 3 e 5 + y–e 0 + x– 1 1/e (2 + e 4)x – 2 – e 4 + 4 e 3 y – 4 e 4 + e 5 z y/e – 1 + z/e = z– 0 = (2 + e 4)x + 4 e 3 y + e 5 z – 1 – 4 e 4 (y + z) / e Observe that since w = f(x, y, z) goes from R 3 to R 2, this linear approximation is really two linear approximations, one for each of the component functions. That is,

Observe that since w = f(x, y, z) goes from R 3 to R 2, this linear approximation is really two linear approximations, one for each of the component functions. That is, The linear approximation for f 1(x, y, z) = x 2 + xy 4 eyz is w 1 = (2 + e 4)x + 4 e 3 y + e 5 z– (1 + 4 e 4). The linear approximation for f 2(x, y, z) = ln(xz+y) is w 2 = (1/e)y + (1/e)z.

Recall that for a function f from Rn to R 1, the derivative matrix is 1 n. In certain situations, it will be convenient to treat this 1 n matrix as a vector. See the definition of a gradient on page 136. Look at Theorem 8 on page 137. Look at Theorem 9 on page 137. Look at the chart at the top of page 138.