Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.6 The Chain Rule - Exercises Set 2.6: 50

Answer

The equation for the tangent line is $y=x.$

Work Step by Step

The equation for the tangent line at $x_0$ is: $$y-f(x_0)=f'(x_0)(x-x_0)$$ Here is $x_0=0$ and $f(x)=\frac{x}{\sqrt{1-x^2}}.$ $$f'(x)=\Big(\frac{x}{\sqrt{1-x^2}}\Big)'=\frac{x'\sqrt{1-x^2}-x(\sqrt{1-x^2})'}{(\sqrt{1-x^2})^2}=\frac{\sqrt{1-x^2}-x\frac{1}{2\sqrt{1-x^2}}(1-x^2)'}{1-x^2}=\frac{1-x^2-x\cdot\frac{(-2x)}{2}}{(1-x^2)^{3/2}}=\frac{1}{(1-x^2)^{3/2}}$$ Now, for $x=0$ we have: $$f(0)=\frac{0}{\sqrt{1-0^2}}=\frac{0}{1}=0$$ $$f'(0)=\frac{1}{(1-0^2)^{3/2}}=\frac{1}{1}=1$$ So, the equation for the tangent line is: $$y-0=1\cdot(x-0)\Rightarrow y=x$$
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.