# Practice%20exercises%202-ANSWERS - 1 Practice exercises for...

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1 Practice exercises for Exam 2 Please remember that if there are homework problems or in-class examples that you have not understood you would be better oﬀ revisiting them ﬁrst. 1. Implicit diﬀerentiation: (a) Diﬀerentiate implicitly e xy = y x Answer: e xy ( y + xy 0 ) = y 0 x - y x 2 (b) Find the equation of the tangent line to the curve y = x - 2 y + 1 through the point where y = 1 Answer: First plug in y = 1 to ﬁnd x at the points on the curve where y = 1. Then implicitly diﬀerentiate to ﬁnd the slope. You’ll end up with a line with slope 1 10 through the point (4 , 1). y = 1 10 x + 3 5 = 0 . 1 x + 0 . 6 2. A parametric curve is described by equations x ( t ) = t 3 + 1 y ( t ) = ln( t - 1) Find y and dy dx when x = 9 in two diﬀerent ways: (a) Directly without determining the cartesian equation for the function y ( x ). Answer: If you know the value of x then you can ﬁnd t . Recall dy dx in terms of dy dt and dx dt . You’ll get 1 / 12 (b) By computing and diﬀerentiating the cartesian equation. Hint: ﬁrst, solve for t from the equation for x , and plug it into the expression for y . Answer: y = ln( 3 x - 1 - 1). Thus y 0 = 1 3 x - 1 - 1 · 1 3 ( x - 1) - 2 / 3 . Setting x = 9 again yields 1 / 12 3. Use linearization to approximate the following: (a) 1 3 . 9 Answer: The expression can be viewed as a function f ( x ) = 1 x = x - 1 / 2 evaluated at 3 . 9. Note that f (4) is much easier to ﬁnd, so instead of f (3 . 9) ﬁnd the tangent line to the graph of f ( x

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Practice%20exercises%202-ANSWERS - 1 Practice exercises for...

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