sin(x+y).Consequently,fx=xcos(x+y).01010.0pointsFind the slope in thex-direction at thepointP(0,2, f(0,2)) on the graph offwhenf(x, y) = 2(2x+y)e−xy.1.slope =-82.slope =-63.slope =-4correct4.slope =-25.slope =-10Explanation:The graph offis a surface in 3-spaceand the slope in thex-direction at the pointP(0,2, f(0,2)) on that surface is the value ofthe partial derivativefxat (0,2). Nowfx= 4e−xy-2(2xy+y2)e−xy.Consequently, atP(0,2, f(0,2))slope =-4.01110.0pointsFind the value offxat (1,3) whenf(x, y) = 3x3-8x2y-2x+ 6y.Correct answer:-41.Explanation:After differentiation,fx=∂f∂x= 9x2-16xy-2.At (1,3), therefore,fxvextendsinglevextendsinglevextendsingle(1,3)=-41.01210.0pointsDetermine∂z∂ywhenz= 4e−x/y.1.∂z∂y=-4y2e−x/y2.∂z∂y=-4xy2e−x/y3.∂z∂y=4y2e−x/y4.∂z∂y=-4xye−x/y5.∂z∂y=4xy2e−x/ycorrect6.∂z∂y=4xye−x/yExplanation:Differentiatingzwith respect toykeepingxfixed we see that∂z∂y= 4e−x/y·∂(-x/y)∂y.

guzman (lg28686) – HW 10 – hanselman – (53860)5Consequently,∂z∂y=4xy2e−x/y.01310.0pointsFindfxwhenf(x, y) =integraldisplayxycos(t4)dt .1.fx= sin(x4)2.fx= 03.fx= cos(x4)correct4.fx= 4x3sin(x4)5.fx= 4x3cos(x4)Explanation:By the Fundamental theorem of calculus,∂f∂x=∂∂xintegraldisplayxycos(t4)dt= cos(x4).01410.0pointsDeterminefyxwhenf(x, y) =x2cosxy .1.fyx=-y2(3 sinxy+xycosxy)2.fyx=-x2(3 sinxy+xycosxy)correct3.fyx= 2y2(3 sinxy-xycosxy)4.fyx=-2y2(3 cosxy+xysinxy)5.fyx=x2(3 cosxy-xysinxy)6.fyx=-2x2(3 cosxy+xysinxy)7.fyx=y2(3 cosxy-xysinxy)8.fyx= 2x2(3 sinxy-xycosxy)Explanation:By the Product Rule,fy=-x3sinxy .But thenfyx=-3x2sinxy-x3ycosxy .Consequently,fxy=-x2(3 sinxy+xycosxy).01510.0pointsFind the value offxx+fyyat (1,-1) whenf(x, y) =1xy-3x2+ 2y2.1.(fxx+fyy)vextendsinglevextendsinglevextendsingle(1,−1)=-72.(fxx+fyy)vextendsinglevextendsinglevextendsingle(1,−1)=-6correct3.(fxx+fyy)vextendsinglevextendsinglevextendsingle(1,−1)=-54.(fxx+fyy)vextendsinglevextendsinglevextendsingle(1,−1)= 35.(fxx+fyy)vextendsinglevextendsinglevextendsingle(1,−1)= 2Explanation:Differentiatingftwice with respect toxweobtain∂f∂x=-1x2y-6x,∂2f∂x2=2x3y-6.Repeating forywe next obtain∂f∂y=-1xy2+ 4y,∂2f∂x2=2xy3+ 4.

guzman (lg28686) – HW 10 – hanselman – (53860)6Thus at (1,-1),(fxx+fyy)vextendsinglevextendsinglevextendsingle(1,−1)=-6.01610.0pointsDeterminedzdtwhenz=xln(x+ 11y)andx= sint ,y= cost .1.dzdt=ln(x+ 11y) sint-11xcostx+ 11y2.dzdt= ln(x+ 11y) cost-11xsintx+ 11y3.dzdt= ln(x+ 11y) cost+x(sint-cost)x+ 11y4.dzdt= ln(x+11y) cost+xcost-11xsintx+ 11ycorrect5.dzdt= ln(x+11y) sint+xsint-11xcostx+ 11yExplanation:By the Chain Rule for Partial Differentia-tion,dzdt=∂z∂xdxdt+∂z∂ydydt.Here, we have that∂z∂x=xx+ 11y+ ln(x+ 11y),dxdt= costand∂z∂y=11xx+ 11y,dydt=-sint .It follows thatdzdt= ln(x+ 11y) cost+xcost-11xsintx+ 11y.keywords:01710.0pointsUse the Chain Rule to find∂z∂twhenz=x2-3xy+y2,andx= 3s-2t ,y=st .1.∂z∂t= 6x-9y-3xt+ 2yt2.∂z∂t=-4x+ 6y-3xs+ 2yscorrect3.∂z∂t= 6x+ 6y-3xt+ 2yt4.∂z∂t=-4x-9y-3xs+ 2ys5.∂z∂t

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