Chemistry 1S - Dr Paul May Calculus Question for January Exam 2010 1. Consider the function: = y(x) (a) Differentiate using the Quotient Rule. y(x) (5 marks) (b) Use the Chain Rule (or Function-of-a-Function Rule) to differentiate: = (x 5)1 v(x) – (5 marks) (c) Using your answer to part (b) and by rewriting as: y(x) = (x + 3)×(x 5)1, y(x) – differentiate using the Product Rule, and check that the answer is the same as y(x) that from part (a). (6 marks) (d) Determine the co-ordinates (x,y) of the stationary point(s) of if it has any. y(x), (6 marks) (e) Using your answer to part (d) sketch between = 2 and = +4. y(x) x x (8 marks) Answers 3) (a) Using Quotient Rule, = , y(x) (b) = (x 5)-1; using F-of-F Rule, dv/dx = 1(x 5)-2(1) = (x 5)-2 = v(x) – – (c) If = (x + 3)×(x 2)-1; using Product Rule: = y(x) y(x) u(x)×v(x), = + and noticing that is the same as that in part (b) so that u v v(x) we’ve already worked out dv/dx, = (x + 3)×() + (x 5)-1×(1) = Putting over a common denominator: = This is the same as (a). (d) At the t.p. dy/dx = 0, so = 0, so the only turning points are at = ± infinity. This x is a bit of a trick question, since the curve doesn’t really have any turning points real as such.
Chemistry 1S - Dr Paul May Calculus Question for January Exam 2010 1. Consider the function: y(x) = (a) Differentiate y(x) using the Quotient Rule. (5 marks) (b) Use the Chain Rule (or Function-of-a-Function Rule) to differentiate: v(x) = (x – 5)1 (5 marks) (c) Using your answer to part (b) and by rewriting y(x) as: y(x) = (x + 3)×(x – 5)1, differentiate y(x) using the Product Rule, and check that the answer is the same as that from part (a). (6 marks) (d) Determine the co-ordinates (x,y) of the stationary point(s) of y(x), if it has any. (6 marks) (e) Using your answer to part (d) sketch y(x) between x = 2 and x = +4. (8 marks) Answers 3) (a) Using Quotient Rule, y(x) = , (b) v(x) = (x 5)-1; using F-of-F Rule, dv/dx = 1(x – 5)-2(1) = (x – 5)-2 = (c) If y(x) = (x + 3)×(x 2)-1; using Product Rule: y(x) = u(x)×v(x), = u + v and noticing that v(x) is the same as that in part (b) so that we’ve already worked out dv/dx, = (x + 3)×() + (x 5)-1×(1) = Putting over a common denominator: = This is the same as (a). (d) At the t.p. dy/dx = 0, so = 0, so the only turning points are at x = ± infinity. This is a bit of a trick question, since the curve doesn’t really have any real turning points as such.
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