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Consider the function f: R²R given by (a) Compute the partial derivatives at the point (1, 0): fx(x, y) = 3 X fy(x, y) = fxx (x, y) = 1 fxy(x, y) = 1 fyx (x, y) = 1 fyy(x, y) = 1 0 2 ✓ ✓ x+ X (d) If x = X X (b) (1, 0) is a local maximum of the function f. (c) The tangent plane to the graph of z = f(x, y) at the point (1, 0, 1) can be described by the equation 1 xy+ z = 0 X f(x, y) = x²y + sin(xy) + 1 X (s²+1²) and y = s - t², then at the point (s, t) = (1, 1), af dt (e) The maximum rate of change of f(x, y) at the point (x, y) = (1, 0) is 1 X - is equal to 0 X

Consider the function f: R²R given by (a) Compute the partial derivatives at the point (1, 0): fx(x, y) = 3 X fy(x, y) = fxx (x, y) = 1 fxy(x, y) = 1 fyx (x, y) = 1 fyy(x, y) = 1 0 2 ✓ ✓ x+ X (d) If x = X X (b) (1, 0) is a local maximum of the function f. (c) The tangent plane to the graph of z = f(x, y) at the point (1, 0, 1) can be described by the equation 1 xy+ z = 0 X f(x, y) = x²y + sin(xy) + 1 X (s²+1²) and y = s - t², then at the point (s, t) = (1, 1), af dt (e) The maximum rate of change of f(x, y) at the point (x, y) = (1, 0) is 1 X - is equal to 0 X

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Need help with part b). Thank you :)

 

Consider the function f : R² → R given by (a) Compute the partial derivatives at the point (1, 0): fx(x, y) fy(x, y) = 2 fxx(x, y) = fxy(x, y) = 1 fyx (x, y) = 1 = 3 X fyy(x, y) : 0 = 1 1 X X X f(x, y) = x²y + sin(xy) + 1 X (b) (1, 0) is a local maximum ♦ of the function f. (c) The tangent plane to the graph of z = f(x, y) at the point (1, 0, 1) can be described by the equation ✔ x+ 1 x y+ z = 0 X (d) If x = 1/(s² + 1²) and y = s - t², then at the point (s, t) = (1, 1), af Ət (e) The maximum rate of change of f(x, y) at the point (x, y) = (1, 0) is 1 is equal to 0 X X
Transcribed Image Text:Consider the function f : R² → R given by (a) Compute the partial derivatives at the point (1, 0): fx(x, y) fy(x, y) = 2 fxx(x, y) = fxy(x, y) = 1 fyx (x, y) = 1 = 3 X fyy(x, y) : 0 = 1 1 X X X f(x, y) = x²y + sin(xy) + 1 X (b) (1, 0) is a local maximum ♦ of the function f. (c) The tangent plane to the graph of z = f(x, y) at the point (1, 0, 1) can be described by the equation ✔ x+ 1 x y+ z = 0 X (d) If x = 1/(s² + 1²) and y = s - t², then at the point (s, t) = (1, 1), af Ət (e) The maximum rate of change of f(x, y) at the point (x, y) = (1, 0) is 1 is equal to 0 X X
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