Partial Derivatives Practice
Find fx and fy for the following multi-variable functions.
- f(x,y) = xy2
- f(x,y) = 2x3y2 + y
- f(x,y) = 1.45x2y - xy + 2x - 3
- f(x,y) = (x/y)
- f(x,y) = xy
- f(x,y) = x ln y
- f(x,y) = 2exy
- f(x,y) = 3xy + x3y2
- f(x,y) = 2x3(1+y2)2
- f(x,y) = (1+x2y)3
Find all second derivatives (fxx, fxy, fyx and fyy) of these functions.
- f(x,y) = 2x5y - x2
- f(x,y) = x + 3y2
- f(x,y) = xey
- f(x,y) = ln(x2 + y2)
- f(x,y) = 2y3x
Answers
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- fx = y2; fy = 2xy
- fx = 6x2y2; fy = 4x3y + 1
- fx = 2.9xy - y + 2; fy = 1.45x2 - x
- fx = 1/y; fy = -x/y2
- fx = yxy-1; fy = xy(ln x)
- fx = ln y; fy = x/y
- fx = 2yexy; fy = 2xexy
- fx = y3xy(ln 3) + 3x2y2; fy = x3xy(ln 3) + 2x3y
- fx = 6x2(1 + y2)2; fy = 8x3y(1 + y2)
- fx = 6xy(1 + x2y)2; fy = 3x2(1 + x2y)2
- fx = 10x4y - 2x; fy = 2x5; fxx = 40x3y - 2; fxy = 10x4; fyx = 10x4; fyy = 0
- fx = 1; fy = 6y; fxx = 0; fxy = 0; fyx = 0; fyy = 6
- fx = ey; fy = xey; fxx = 0; fxy = ey; fyx = ey; fyy = xey
- fx = 2x/(x2 + y2); fy = 2y/(x2 + y2); fxx = (2y2 - 2x2)/(x2 + y2)2; fxy = -4xy/(x2 + y2)2; fyx = -4xy/(x2 + y2)2; fyy = (2x2 - 2y2)/(x2 + y2)2
- fx = 2y3x(ln 3); fy = 2y3x(ln 2); fxx = 2y3x(ln 3)2; fxy = 2y3x(ln 3)(ln 2); fyx = 2y3x(ln 3)(ln 2); fyy = 2y3x(ln 2)2
Errors corrected 9/22/06 - sas