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 - Page 188: 35



Work Step by Step

We have our original function below: $f(x)=\frac{1}{1+x^2}$ We see that we have a quotient, so let's remember the quotient rule: $ \frac{d}{dx}[\frac{f(x)}{g(x)}]=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2} $ Using this, let's find the derivative of both the numerator and denominator of $f(x)$: Numerator = $1$ Derivative of Numerator = $0$ Denominator = $1+x^2$ Derivative of Denominator = $2x$ Now, let's plug that into the quotient rule: $f'(x)=\frac{0(1+x^2)-2x(1)}{(1+x^2)^2}$ And simplify: $f'(x)=\frac{-2x}{(1+x^2)^2}$ Now we have the derivative. Let's now use the point $(-1,\frac{1}{2})$ to find the slope. To do that, we need to plug in $x=1$ for all $x$ in $f'(x)$: $f'(-1)=\frac{-2(-1)}{(1+(-1)^2)^2} = \frac{2}{4} = \frac{1}{2}$ So now we know that the slope ($m$) equals $\frac{1}{2}$ So we can use the point-slope form of a line to now get an equation: $y-y_1=m(x-x_1)$ Now we plug in our points. Remember from the point $(-1,\frac{1}{2})$, $x=-1$ and $y=\frac{1}{2}$ $y-\frac{1}{2}=\frac{1}{2}(x+1)$ Now we just have to solve for $y$: $y-\frac{1}{2}=\frac{1}{2}x+\frac{1}{2}$ $y=\frac{1}{2}x+1$ And that's our equation for the tangent line of $f(x)=\frac{1}{1+x^2}$ through the point $(-1,\frac{1}{2})$. To get a graph, just use a graphing utility and graph the original function and the tangent line.
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