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Extra info for Calculus 2c-2, Examples of Description of Surfaces Partial Derivatives, Gradient, Directional Derivative and Taylor's Formula

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All these examples are very simple because they should train the reader to use a new method. g. 6, where the chain rule is the most efficient one. 2 we must admit that the chain rule is more difficult to apply. 3 It can be proved that the differential equation dw = w 2 + u2 , du u ∈ R, among its solutions has w = X(u), u ∈ R, where X(0) = 1, u ∈ R, where Y (0) = 2. and w = Y (u), Let F (u) = f (X(u), Y (u)), f (x, y) = ln(1 + xy 2 ). Find the derivative F (0). com 37 Calculus 2c-2 The chain rule A The chain rule.

3 √ 3, so f 1 (1, 2, 3); √ (1, 1, 1) 3 √ 1 12 = √ · 2[x + y + z](x,y,z)=(1,2,3) = √ = 4 3. 2 Find in each of the following cases the directional derivative of the function f at the point given by the index 0 in the direction of the point given by the index 1. z x y + + defined for xyz = 0 from (x0 , y0 , z0 ) = (1, −1, 1) to y z x (x1 , y1 , z1 ) = (3, 1, 2). 1) f (x, y, z) = xyz + 2) f (x, y, z) = 2x3 y − 3y 2 z defined in R3 from (x0 , y0 , z0 ) = (1, 2, −1) to (x1 , y1 , z1 ) = (3, −1, 5). 3) f (x, y, z) = x ln(1 + eyz ) defined in R3 from (x0 , y0 , z0 ) = (1, 1, 0) to (x1 , y1 , z1 ) = (0, 0, −1).

We get by insertion, 1 + u2 + v 2 · cos(Arcsin u) = + F (u, v) = 1 + u2 + v 2 · 1 − u2 , hence ⎧ √ √ ⎨ u 1 − u2 ∂F u 1 + u2 + v 2 √ =√ − =u ⎩ ∂u 1 + u2 + v 2 1 − u2 1 − u2 − 1 + u2 + v 2 ⎫ 1 + u2 + v 2 ⎬ , 1 − u2 ⎭ and 1 − u2 . 1 + u2 + v 2 ∂F =v ∂v The results agree. ♦ 5) When f (x, y) = x sinh y and (x, y) = X(u, v) = u3 v, ln v + ln v), (u, v) ∈ R2+ , then the composition of the functions is defined and of class C ∞ . From ∂f = x cosh y, ∂y ∂f = sinh y, ∂x and ∂x = 3u2 v, ∂u 1 ∂y = , ∂u u ∂x = u3 , ∂v 1 ∂y = , ∂v v follows that ∂F 1 = sinh y · 3u2 v + x · cosh y · , ∂u u and ∂F 1 = sinh y · u3 + x · cosh y · .

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