Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.1 Definition of the Derivative - Exercises - Page 103: 45


$ f'(0)=0$, Equation of the tangent line is: $ y=1$

Work Step by Step

Recall that $ f'(a)=\lim\limits_{h \to 0}\frac{f(a+h)-f(a)}{h}$ Given that $ a=0$ and $ f(x)=\frac{1}{x^{2}+1}$. Therefore, $ f'(0)=\lim\limits_{h \to 0}\frac{f(0+h)-f(0)}{h}=\lim\limits_{h \to 0}\frac{\frac{1}{h^{2}+1}-1}{h}$ $=\lim\limits_{h \to 0}\frac{1-h^{2}-1}{h(h^{2}+1)}=\lim\limits_{h \to 0}\frac{-h}{h^{2}+1}=0$ Equation of the tangent line is of the form $ y-f(a)=f'(a)(x-a)$ Knowing that $ f(a)=f(0)=1, f'(a)=f'(0)=0$ and $ a=0$, we obtain the equation of the tangent line through $ a $ as below. $ y-1=0(x-0)$ Or $ y=1$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.