Explanation Given: The function, g ( x ) = x e x . Concept Used: The quotient rule of derivatives: d d x [ f ( x ) g ( x ) ] = g ( x ) ⋅ d d x f ( x ) − f ( x ) ⋅ d d x g ( x ) ( g ( x ) ) 2 Power rule of derivatives: d d x ( x n ) = n x n − 1 d d x ( e x ) = e x Calculation: In order to find thenth derivative of the given function, use the quotient rule of derivatives. Observe that the given function is a quotient of two functions x and ( e x ) . So, using the product rule and power rule of derivatives, first derivative can be found as g ' ( x ) = d d x ( x e x ) = ( e x ) ⋅ d d x ( x ) − x ⋅ d d x ( e x ) ( e x ) 2 = e x ⋅ ( 1 ) − x ⋅ ( e x ) ( e x ) 2 = e x ( 1 − x ) ( e x ) 2 = 1 − x e x = − ( x − 1 ) e x So, the first derivative is g ' ( x ) = − ( x − 1 ) e x . Again, using the product rule of derivatives, differentiating above first derivative with respect to x , it gives g ' ' ( x ) = d d x ( 1 − x e x ) = ( e x ) ⋅ d d x ( 1 − x ) − ( 1 − x ) ⋅ d d x ( e x ) ( e x ) 2

Bundle: Biocalculus: Calculus, Probability, And Statistics For The Life Sciences + Webassign Printed Access Card For Stewart/day's Biocalculus: Calculus For Life Sciences, 1st Edition, Multi-term

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ISBN: 9781305779037

Author: James Stewart, Troy Day

Publisher: Cengage Learning

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Chapter 3.4, Problem 44E

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Thenth derivatives g(n)(x) for the function, g(x)=xex .

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Chapter 3.4, Problem 43E

Chapter 3.4, Problem 45E

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3. Differentiate the following functions. Show your work where applicable. a) y = e³x b) f(x)=2 cos(5x) c) y = 1 - 2 d) y = In|secx| e) f(t) = t² e√t f) f(x) = 1+x x sin x 3

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Thenth derivatives g ( n ) ( x ) for the function, g ( x ) = x e x .

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