ha (lvh262) – Homework 14.3 – karakurt – (56295) 1This print-out should have 29 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsDetermine fx− fywhenf(x, y) = 2x2− 3xy + 2y2− x + 2y .1. fx− fy= x + y − 32. fx− fy= 7x − 7y + 13. fx− fy= 7x − 7y − 3 correct4. fx− fy= x + y + 15. fx− fy= x − 7y + 16. fx− fy= 7x + y − 3Explanation:After differentiation we see t hatfx= 4x − 3y − 1 , fy= −3x + 4y + 2 .Consequently,fx− fy= 7x − 7y − 3 .002 10.0 pointsDetermine fxwhenf(x , y) = (x2− 2y)(y2− x) .1. fx= 2y − 2xy2− 3x22. fx= 2xy2+ 2y − 3x2correct3. fx= y − 4xy2+ 3x24. fx= 2xy2− 2y − 3x25. fx= 4xy2− y + 3x26. fx= y + 4xy2+ 3x2Explanation:From the Product Rule we see thatfx= 2x(y2− x) − (x2− 2y) .Consequently,fx= 2xy2+ 2y − 3x2.003 10.0 pointsDetermine fxwhenf(x, y) = x sin(y − x) − cos(y − x) .1. fx= −2 sin(y − x) − x cos(y − x)2. fx= −x sin(y − x)3. fx= x cos(y − x)4. fx= x sin(y − x)5. fx= −x cos(y − x) correct6. fx= − cos(y − x) − x sin(y − x)7. fx= 2 sin(y − x) − x cos(y − x)8. fx= x cos(y − x) − sin(y − x)Explanation:From the Product Rule we see thatfx= sin(y − x) − x cos(y − x) − sin(y − x) .Consequently,fx= −x cos(y − x) .004 10.0 pointsDetermine fxwhenf(x, y) = x sin(x + 2y) + cos(x + 2y) .ha (lvh262) – Homework 14.3 – karakurt – (56295) 21. fx= 2 cos(x + 2y) + x sin(x + 2y)2. fx= 2 si n(x + 2y) + x cos(x + 2y)3. fx= x cos(x + 2y) correct4. fx= −2x sin(x + 2y)5. fx= 2x cos(x + 2y)6. fx= 2 si n(x + 2y) − x cos(x + 2y)7. fx= 2 cos(x + 2y) − x sin(x + 2y)8. fx= −x sin(x + 2y)Explanation:From the Product Rule we see thatfx= sin(x+2y)+x cos( x+2 y)− sin(x+2y) .Consequently,fx= x cos(x + 2y) .005 10.0 pointsDetermine fywhenf(x, y) = y cos(3x − y) − sin(3x − y) .1. fy= 2 cos(3x−y)+y sin(3x−y) correct2. fy= −y cos(3x − y)3. fy= 2 si n(3x − y) − y cos(3 x − y)4. fy= −y sin(3x − y)5. fy= −2 sin(3x − y) + y cos (3x − y)6. fy= −2 cos(3x − y) − y sin (3x − y)7. fy= y sin(3x − y)8. fy= y cos(3x − y)Explanation:From the Product Rule we see thatfy= cos(3x− y) + y sin(3x− y) + cos(3x− y) .Consequently,fy= 2 cos(3x − y) + y sin(3x − y) .006 10.0 pointsFind the slope in the x-direction at thepoint P (0, 2, f(0, 2)) on the graph of f whenf(x, y) = 4(y2− x2) l n( x + y) .1. slope = 62. slope = 43. slope = 24. slope = 105. slope = 8 correctExplanation:The graph of f is a surface in 3-spaceand the slope in the x-direction at the pointP (0, 2, f(0, 2)) on that surface is the value ofthe partial derivative fxat (0, 2). Nowfx= 4−2x ln(x + y) +y2− x2x + y.Consequently, at P (0, 2, f(0, 2))slope = 2 × 4 = 8 .007 10.0 pointsDetermine fywhenf(x , y) =x − 2 y2 x + y.1. fy=3 x(2 x + y)22. fy= −4 x(2 x + y)2ha (lvh262) – Homework 14.3 – karakurt – (56295) 33. fy=5 x(2 x + y)24. fy= −5 x(2 x + y)2correct5. fy=4 x(2 x + y)26. fy= −3 x(2 x + y)2Explanation:From the Quotient Rule we see thatfy=−2 (2 x + y) − ( x − 2 y)(2 x + y)2.Consequently,fy= −5 x(2 x + y)2.008 10.0 pointsFind the value of fxat (3, 3) whenf(x, y) = 6x3− 3x2y − 6x + 3y.Correct answer: 102.Explanation:After differentiation,fx=∂f∂x= 18x2− 6xy − 6.At (3 , 3), therefore,fx(3,3)= 102.009 10.0 pointsDetermine∂z∂ywhen z = 3ex/2y.1.∂z∂y=3x2yex/2y2.∂z∂y= −3x2yex/2y3.∂z∂y= −3x2y2ex/2ycorrect4.∂z∂y= −32y2ex/2y5.∂z∂y=32y2ex/2y6.∂z∂y=3x2y2ex/2yExplanation:Differentiating z with respect to y keepingx fixed we see that∂z∂y= 3ex/2y·∂(x /2y)∂y.Consequently,∂z∂y= −3x2y2ex/2y.010 10.0 pointsDetermine h = h(x, y) so that∂f∂x=h(x, y)(3 x2+ 3 y2)2whenf(x, y) =4x2y3 x2+ 3 y2.1. h(x, y) = 2 4xy22. h(x, y) = 1 2xy23. h(x, y) = 2 4x3y4. h(x, y) = 2 4xy3correct5. h(x, y) = 1 2xy3ha (lvh262) – Homework 14.3 – karakurt – (56295) 46. h(x, y) = 12x3yExplanation:Differentiating with respect to x using thequotient rule we obtai n∂f∂x=8xy(3x2+ 3 y2) − 24x3y(3x2+ 3y2)2.Consequently,h(x, y) = 24xy3.011 10.0 pointsFind the value of fxat (2, 1) whenf(x, y) =2x2+ 3xyx + y.Correct answer: 2.11111.Explanation:After differentiation using the quotient rulewe see thatfx=(x + y) (4x + 3y) − 2x2− 3xy(x + y)2=2x2+ 4xy + 3y2(x + y)2.At (2 , 1), therefore.fx(2, 1)=199.012 10.0 pointsFind fxwhenf(x, y) = ln(px2+ y2− x) .1. fx=xpx2+ y22. fx=1px2− y23. fx= −1px2+ y2correct4. fx=xpx2− y25. fx= −1px2− y26. fx=1px2+ y2Explanation:Differentiating with respect to x keeping yfixed, we see thatfx=xpx2+ y2− 1px2+ y2− x.Butxpx2+ y2− 1 =x −px2+ y2px2+ y2.Consequently,fx= −1px2+ y2.013 10.0 pointsFind fxwhenf(x, y) =Zxycost4dt .1. fx= sinx42. fx= cosx4correct3. fx= 04. fx= 4x3sinx45. fx= 4x3cosx4Explanation:ha (lvh262) – Homework 14.3 – karakurt – (56295) 5By the Fundamental theorem of calculus,∂f∂x=∂∂xZxycost4dt= cosx4.014 10.0 pointsDetermine fxywhenf(x, y) =12y tan−1(xy) .1. fxy= −xy2(1 + x2y2)2. fxy=x(1 + x2y2)23. fxy=xy2(1 + x2y2)4. fxy= −y2(1 + x2y2)5. fxy= −x(1 + x2y2)26. fxy=y(1 + x2y2)2correctExplanation:Since we can choo se whether to differentiatewith respect to x or y first, for simplicity wewill choose to differentiate first with respectto x because then the algebra is simpler.Indeed, by the Chain Rule,fx=12y∂∂xtan−1(xy)=12y211 + (xy)2=y22(1 + x2y2).Thus by the Quotient Rulefxy=122y(1 + x2y2) − y2(2x2y)(1 + x2y2)2.Consequently,fxy=y(1 + x2y2)2.keywords: partial differentiation, mixed par-tial derivative, Chain Rule, inverse tangent,PartialDiffMV, PartialDiffMVExam,015 10.0 pointsDetermine fxywhenf(x, y) = y tan−1(xy) .1. fxy=2x(1 + x2y2)22. fxy=2x1 + x2y23. fxy=x(1 + x2y2)24. fxy=2y(1 + x2y2)2correct5. fxy=y(1 + x2y2)26. fxy=2y1 + x2y2Explanation:Differentiating f partia lly with respect tox, using also the Chain Rule andddxtan−1x =11 + x2,we see thatfx= yy1 + x2y2.But after simplification,fx=y2(1 + x2y2).Differentiating partially now with respect toy we see thatfxy=(1 + x2y2) · 2y − y2(2x2y)(1 + x2y2)2.ha (lvh262) – Homework 14.3 – karakurt – (56295) 6But after simplification once again,fxy=2y(1 + x2y2)2.keywords: partial differentiation, mixed par-tial derivative, Chain Rule, inverse tangent,016 10.0 pointsFind utwhenu = xe−6tsin θ .1. ut= −6xe−6tsin θ