Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises: 32

Answer

The equation of the tangent line $l$ to the given curve at $(0,\frac{1}{2})$ is $$(l): y=\frac{1}{4}x+\frac{1}{2}$$

Work Step by Step

$$y=f(x)=\frac{1+x}{1+e^x}$$ 1) Find $f'(x)$ $$f'(x)=\frac{(1+x)'(1+e^x)-(1+x)(1+e^x)'}{(1+e^x)^2}$$ $$f'(x)=\frac{1\times(1+e^x)-e^x(1+x)}{(1+e^x)^2}$$ $$f'(x)=\frac{1+e^x-e^x-e^xx}{(1+e^x)^2}$$ $$f'(x)=\frac{1-e^xx}{(1+e^x)^2}$$ 2) The slope of the tangent line $l$ to the given curve at point $A(a,b)$ is $f'(a)$. Therefore, at $(0,\frac{1}{2})$, the slope of tangent line $l$ is $$f'(0)=\frac{1-e^0\times0}{(1+e^0)^2}$$ $$f'(0)=\frac{1-0}{(1+1)^2}=\frac{1}{4}$$ The equation of the tangent line $l$ to the given curve at $(0,\frac{1}{2})$ has the following form: $$(l): (y-\frac{1}{2})=f'(0)(x-0)$$ $$(l): y=\frac{1}{4}x+\frac{1}{2}$$
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