Answered: (a) What is the slope of the line…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter5: Inverse, Exponential, And Logarithmic Functions

Section: Chapter Questions

Problem 18T

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Question

(a) What is the slope of the line tangent to y = e^-x/(1 + e^-x) at x = 0?

(b) Write the equation of the line tangent to the graph of y = e^-x/(1 + e^-x) at x = 0.

Expert Solution

Step 1

Given- $y = \frac{e^{- x}}{(1 + e^{- x})}$

To find - The slope of the line tangent to above curve at x=0.

Concept Used- The slope of line is same as the slope of the curve at the given point and to find the slope, we have to differentiate both sides with respect to x.,

Formula used- The quotient formula of derivative which is given below,

$\frac{d}{d x} (\frac{u}{v}) = \frac{v \cdot u' - u \cdot v'}{v^{2}}$

Step 2

Explanation- Rewrite the given expression,

$y = \frac{e^{- x}}{(1 + e^{- x})}$

Now, to find the slope, differntiate both sides with respect to x, we get,

$\frac{d y}{d x} = \frac{(1 + e^{- x}) \cdot (- e^{- x}) - (- e^{- x}) \cdot e^{- x}}{{(1 + e^{- x})}^{2}} = \frac{- e^{- x} - e^{- 2 x} + \cdot e^{- 2 x}}{{(1 + e^{- x})}^{2}} = \frac{- e^{- x}}{{(1 + e^{- x})}^{2}}$

As we have to find the slope at x=0, so solving further,

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(a) What is the slope of the line tangent to y = e-x/(1 + e-x) at x = 0? (b) Write the equation of the line tangent to the graph of y = e-x/(1 + e-x) at x = 0.