ha lvh262 Homework 14 3 karakurt 56295 This print out should have 29 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 10 0 points Determine fx fy when 2 2 f x y 2x 3xy 2y x 2y 1 fx fy x y 3 2 fx fy 7x 7y 1 3 fx fy 7x 7y 3 correct 1 Explanation From the Product Rule we see that fx 2x y 2 x x2 2y Consequently fx 2xy 2 2y 3x2 003 10 0 points Determine fx when f x y x sin y x cos y x 4 fx fy x y 1 1 fx 2 sin y x x cos y x 5 fx fy x 7y 1 2 fx x sin y x 6 fx fy 7x y 3 Explanation After differentiation we see that fx 4x 3y 1 fy 3x 4y 2 Consequently fx fy 7x 7y 3 3 fx x cos y x 4 fx x sin y x 5 fx x cos y x correct 6 fx cos y x x sin y x 7 fx 2 sin y x x cos y x 002 10 0 points Determine fx when f x y x2 2y y 2 x 1 fx 2y 2xy 2 3x2 2 fx 2xy 2 2y 3x2 correct 8 fx x cos y x sin y x Explanation From the Product Rule we see that fx sin y x x cos y x sin y x Consequently fx x cos y x 3 fx y 4xy 2 3x2 4 fx 2xy 2 2y 3x2 004 10 0 points 5 fx 4xy 2 y 3x2 Determine fx when 6 fx y 4xy 2 3x2 f x y x sin x 2y cos x 2y ha lvh262 Homework 14 3 karakurt 56295 2 From the Product Rule we see that 1 fx 2 cos x 2y x sin x 2y 2 fx 2 sin x 2y x cos x 2y 3 fx x cos x 2y correct fy cos 3x y y sin 3x y cos 3x y Consequently fy 2 cos 3x y y sin 3x y 4 fx 2x sin x 2y 006 5 fx 2x cos x 2y 10 0 points 6 fx 2 sin x 2y x cos x 2y Find the slope in the x direction at the point P 0 2 f 0 2 on the graph of f when 7 fx 2 cos x 2y x sin x 2y f x y 4 y 2 x2 ln x y 8 fx x sin x 2y Explanation From the Product Rule we see that fx sin x 2y x cos x 2y sin x 2y Consequently 2 slope 4 3 slope 2 4 slope 10 fx x cos x 2y 005 1 slope 6 10 0 points Determine fy when f x y y cos 3x y sin 3x y 1 fy 2 cos 3x y y sin 3x y correct 5 slope 8 correct Explanation The graph of f is a surface in 3 space and the slope in the x direction at the point P 0 2 f 0 2 on that surface is the value of the partial derivative fx at 0 2 Now y 2 x2 fx 4 2x ln x y x y Consequently at P 0 2 f 0 2 2 fy y cos 3x y slope 2 4 8 3 fy 2 sin 3x y y cos 3x y 007 4 fy y sin 3x y 5 fy 2 sin 3x y y cos 3x y Determine fy when f x y 6 fy 2 cos 3x y y sin 3x y 7 fy y sin 3x y 8 fy y cos 3x y Explanation 10 0 points 1 fy 3x 2 x y 2 2 fy 4x 2 x y 2 x 2y 2x y ha lvh262 Homework 14 3 karakurt 56295 3 fy 5x 2 x y 2 4 fy 5 fy 5x correct 2 x y 2 4x 2 x y 2 6 fy 3x 2 x y 2 Explanation From the Quotient Rule we see that fy 2 2 x y x 2 y 2 x y 2 2 z 3x ex 2y y 2y 3 z 3x 2 ex 2y correct y 2y 4 3 z 2 ex 2y y 2y 5 z 3 x 2y e y 2y 2 6 z 3x x 2y e y 2y 2 Explanation Differentiating z with respect to y keeping x fixed we see that Consequently fy 008 5x 2 x y 2 x 2y z 3ex 2y y y Consequently 10 0 points z 3x 2 ex 2y y 2y Find the value of fx at 3 3 when f x y 6x3 3x2 y 6x 3y 010 f h x y x 3 x2 3 y 2 2 Explanation After differentiation f 18x2 6xy 6 x 009 102 1 h x y 24xy 2 10 0 points 2 h x y 12xy 2 3 3 Determine when z 3ex 2y 1 when f x y At 3 3 therefore fx 10 0 points Determine h h x y so that Correct answer 102 fx z 3x x 2y e y 2y 3 z y 4x2 y 3 x2 3 y 2 3 h x y 24x3 y 4 h x y 24xy 3 correct 5 h x y 12xy 3 ha lvh262 Homework 14 3 karakurt 56295 6 h x y 12x3 y Explanation Differentiating with respect to x using the quotient rule we obtain f 8xy 3x2 3 y 2 24x3 y x 3x2 3y 2 2 Consequently h x y 24xy 3 011 3 fx p 4 fx p Correct answer 2 11111 Explanation After differentiation using the quotient rule we see that fx fx But x2 y 2 1 x p 1 x2 y 2 p x2 y 2 x p x x2 y 2 p p 1 x2 y 2 x2 y 2 Consequently fx p 2x2 4xy 3y 2 x y 2 013 2 1 19 9 x cos t4 dt 2 fx cos x4 correct p f x y ln x2 y 2 x 2 fx p Z y 10 0 points Find fx when 1 fx p x2 y 2 10 0 points f x y 1 fx sin x4 012 1 Find fx when At 2 1 therefore fx 1 x x y 4x 3y 2x2 3xy x y 2 correct Explanation Differentiating with respect to x keeping y fixed we see that Find the value of fx at 2 1 when 2x2 3xy x y x2 y 2 x x2 y 2 10 0 points f x y 1 x2 y 2 5 fx p 6 fx p 4 x x2 y 2 1 x2 y 2 3 fx 0 4 fx 4x3 sin x4 5 fx 4x3 cos x4 Explanation ha lvh262 Homework 14 3 karakurt 56295 By the Fundamental theorem of calculus Z x f cos t4 dt x x y cos x4 014 10 0 points 5 keywords partial differentiation mixed partial derivative Chain Rule …