Explanation Given Information: The provided trigonometric function is y = e x cos x over the interval [ 0 , 2 π ] . Formula used: Cosine derivative Rule: d d x [ cos u ] = − sin u d u d x Product rule of differentiation : d d x [ a ( x ) b ( x ) ] = a ( x ) b ' ( x ) + b ( x ) a ' ( x ) Exponential rule of differentiation: d d x [ e u ] = e u d u d x General power derivative rule: d d x x n = n x n − 1 Condition for critical points: d y d x = f ′ ( x ) = 0 If function f ( x ) is increasing, then f ′ ( x ) > 0 and if function is decreasing, then f ′ ( x ) < 0 . From first derivative Test, if f ′ ( x ) changes positive to negative, then function f ( x ) has relative maxima and, if f ′ ( x ) changes negative to positive then function f ( x ) has relative minima. In order words, if function f ( x ) changes from increasing function to decreasing at any point critical, then function f ( x ) will have relative maxima at that point, and if function f ( x ) changes from decreasing function to increasing at any critical point, then function f ( x ) will have relative minima at that point. Calculation: Consider the provided trigonometric function: y = e x cos x Apply the product rule of differentiation to find the derivative of the above function. d y d x = e x d d x [ cos x ] + cos x d d x [ e x ] Differentiate the above function using general power rule, cosine and exponential differentiation rules. d y d x = e x [ − sin x ] + cos x [ e x ] y ′ ( x ) = − e x sin x + e x cos x = e x cos x − e x sin x Apply d y d x = 0 in order to find critical points. e x cos x − e x sin x = 0 e x ( cos x − sin x ) = 0 Further solve. cos x − sin x = 0 − sin x = − cos x sin x cos x = 1 tan x = 1 And, e x ≠ 0 Since, the value of the tan θ is zero at θ = π 4 , and θ = 5 π 4 . So, tan x = 1 tan x = tan π 4 x = π 4 And, tan x = 1 tan x = tan 5 π 4 x = 5 π 4 Thus, critical points are x = π 4 and x = 5 π 4 . Now, rewrite the first derivative of the function. y ′ ( x ) = e x cos x − e x sin x At x = π 6 . y ′ ( π 6 ) = e π 6 cos π 6 − e π 6 sin π 6 = 0.61788 = + v e > 0 At x = π 2 . y ′ ( π 2 ) = e π 2 cos π 2 − e π 2 sin π 2 = − 4.8104 = − v e < 0 At x = 3 π 2 . y ′ ( 3 π 2 ) = e 3 π 2 cos 3 π 2 − e 3 π 2 sin 3 π 2 = 111

Calculus: An Applied Approach (MindTap Course List)

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ISBN: 9781305860919

Author: Ron Larson

Publisher: Cengage Learning

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Chapter 8.4, Problem 61E

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To calculate: The relative extrema of the trigonometric function y=excosx over the interval [0,2π].

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Chapter 8.4, Problem 60E

Chapter 8.4, Problem 62E

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Calculus: An Applied Approach (MindTap Course List)

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The relative extrema of the trigonometric function y = e x cos x over the interval [ 0 , 2 π ] .

Solution Summary: The author calculates the relative extrema of the trigonometric function y=exmathrmcos.

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To calculate: The relative extrema of the trigonometric function y=excosx over the interval [0,2π].

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Chapter 8 Solutions