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of (d) If x = ;(s2 + 12) and y = s - 12, then at the point (s, t) = (1, 1), д is equal to 0 х.

of (d) If x = ;(s2 + 12) and y = s - 12, then at the point (s, t) = (1, 1), д is equal to 0 х.

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 d). 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
Expert Solution
Step 1

Given : fx , y=x2y+sin(xy)+1   x=12s2+t2      ,   y=s-t2

To find : ∂f∂t at 1 , 1

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