Explanation Given: The function, y = 1 s + k e s . 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 the derivative of the given function, use the quotient rule of derivatives. Observe that the given function is a quotient of two functions 1 and ( s + k e s )

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

1st Edition

ISBN: 9781305779037

Author: James Stewart, Troy Day

Publisher: Cengage Learning

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Chapter 3.4, Problem 22E

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The derivative of the function, y=1s+kes .

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Chapter 3.4, Problem 21E

Chapter 3.4, Problem 23E

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Taylor Series Approximation Example- H.W More terms used implies better approximation f(x) 4 f(x) Zero order f(x + 1) = f(x;) First order f(x; + 1) = f(x;) + f'(x;)h 1.0 Second order 0.5 True f(x + 1) = f(x) + f'(x)h + ƒ"(x;) h2 2! f(x+1) 0 x; = 0 x+1 = 1 x h f(x)=0.1x4-0.15x³- 0.5x2 -0.25x + 1.2 51 Taylor Series Approximation H.w: Smaller step size implies smaller error Errors f(x) + f(x,) Zero order f(x,+ 1) = f(x) First order 1.0 0.5 Reduced step size Second order True f(x + 1) = f(x) + f'(x)h f(x; + 1) = f(x) + f'(x)h + "(xi) h2 f(x,+1) O x₁ = 0 x+1=1 Using Taylor Series Expansion estimate f(1.35) with x0 =0.75 with 5 iterations (or & s= 5%) for f(x)=0.1x 0.15x³-0.5x²- 0.25x + 1.2 52

Chapter 3 Solutions

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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The derivative of the function, y = 1 s + k e s .

Solution Summary: The author explains how to find the derivative of the given function using the quotient rule and power rule of derivatives.

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The derivative of the function, y=1s+kes .

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Chapter 3 Solutions