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Name Math 231 April 6, 2017 1. Use the definition of the derivative to find the instantaneous rate of change of the function f (x) = 5x at the point x = 2. Show all your work below. ′ = lim f (2) x →2 5 x - 25 x-2 x 5 x - 25 x-2 1.95 38.66 1.99 39.91 2 ------ 2.01 40.56 2.05 41.90 It appears that the derivative at x = 2 is approximately 40.2 You may need to move in closer. 2. Answer any ten (or more) of the following: a. Is the graph of f(x) = x3 – 4x2 + 5x + 3 increasing or decreasing when x = 2? Give the (calculus) reason for your answer. f ′(x) = 3x 2 − 8x + 5 and f ′(2) = 12 − 16 + 5 = 1 > 0 . Since the derivative is positive when x = 2, the function is increasing there. 3 b. Write the derivative of f (x) = ( sin 5x − cos 7x ) . 2 f ′(x) = 3 ( sin 5x − cos 7x ) ⋅ ( 5 cos 5x + 7 sin 7x ) Math 231 ( April 6, 2017 ) c. Find the instantaneous rate of change of f (x) = ln x 3 x 2 − 4x + 8 at x = 3. 1 x 2 - 4x + 8 = 3ln x + ln ( x 2 - 4x + 8 ) 2 3 1 2x - 4 ′ = + f (x) x 2 x 2 - 4x + 8 12 1 f ′ ( 3 ) = 1+ = 1+ = 1.2 25 5 ( ) f(x) = ln x 3 2 ( ) 7 d. Write the derivative of g(x) = e 4x 3x 2 + 1 . (No need to simplify.) 6 g ′(x) = e 4x ⋅ 7 ( 3x 2 + 1) ⋅ 6x + e 4x ⋅ 8x ( 3x 2 + 1) 2 2 7 e. Find the instantaneous rate of change of y at the point (1, 2), where y is determined by the equation x 2 − 2x + y3 = 7 . 2x − 2 + 3y 2 dy =0 dx dy 2 − 2x 2 − 2 = = =0 dx 12 3y 2 f. Find a second point on the line that is tangent to y = The slope of the tangent is given by y ′ = 4 ⋅ Then, when x = 1, the slope is 4 ⋅ at the point (1, 8). 4x − 5 2 2x 2 − 5x + 7 4 −5 2 2−5+ 7 . = −1 . Other points on the tangent line include (2, 7), (5, 4), (0, 9), (!3, 12), etc. The equation of the tangent line is y = ! (x ! 1) + 8 or y = 9 ! x. Math 231 April 6, 2017 g. At what points is the tangent to y = 2x 3 − 3x 2 − 36x + 10 horizontal? A horizontal line has a slope of zero. Therefore, y′ = 6x 2 − 6x − 36 = 6 ( x 2 − x − 6 ) = 6 ( x − 3 )( x + 2 ) = 0 The tangent is horizontal when x = 3 or x = !2, at the points (3, !71) and (!2, 54). h. Use the derivative rules for sin x and cos x to show that cot x = d (cot x) = − csc 2 x . dx cos x sin x 2 2 d sin x ⋅ -sin x - cos x ⋅ cos x -sin 2 x - cos 2 x - ( sin x + cos x ) -1 cot x = = = = = -csc 2 x 2 2 2 2 dx sin x sin x sin x sin x i. Use the definition to prove that d [ kf (x)] dx j. d[kf (x)] = kf ′(x) (where k is a constant). dx k ( f (x + h) − f (x) ) ( f (x + h) − f (x) ) kf (x + h) − kf (x) = lim = k lim = kf ′(x) h →0 h →0 h →0 h h h = lim Compare the slope of f(x) = x + f ′(x) = 1 − 2 x 2 at the point (1, 3). f ′(1) = 1 − 2 = −1 ; g ′(x) = 2 to that of g(x) = x 3 2 x : g ′(1) = 3 2 The curves are almost, but not quite, perpendicular at (1, 3). Math 231 k. Find p′(0) for p(x) = p′(x) = p′(0) = l. 4x − 3 . 7x + 10 ( 7x + 10 ) ⋅ 4 − ( 4x − 3 ) ⋅ 7 28x + 40 − 28x + 21 61 = = 2 2 2 ( 7x + 10 ) ( 7x + 10 ) ( 7x + 10 ) 61 100 Write the equation of the tangent line to the graph of y = tan −1 x . 1 1 ′ = . f (1) 2 1+ x 2 1 1 The tangent line equation is y = ( x - 1) + .785 = x + .285 2 2 ′ = The slope of the tangent line is given by f (x) April 6, 2017