• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q. 25

Expert-verified
Found in: Page 210

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Find the derivatives of the functions in Exercises 21–46. Keep in mind that it may be convenient to do some preliminary algebra before differentiating. $f\left(x\right)=\frac{1}{\sqrt{{x}^{2}+1}}$

See the step by step solution

## Step 1. Given Information

The given function is $f\left(x\right)=\frac{1}{\sqrt{{x}^{2}+1}}$

## Step 2. Calculation

Differentiate both the sides with respect to x, we get,

$f\text{'}\left(x\right)=\frac{d}{dx}{\left({x}^{2}+1\right)}^{-\frac{1}{2}}\phantom{\rule{0ex}{0ex}}=-\frac{1}{2}{\left({x}^{2}+1\right)}^{-\frac{1}{2}-1}\frac{d}{dx}\left({x}^{2}+1\right)\phantom{\rule{0ex}{0ex}}=-\frac{1}{2}{\left({x}^{2}+1\right)}^{-\frac{3}{2}}\left(\frac{d}{dx}\left({x}^{2}\right)+\frac{d}{dx}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=-\frac{1}{2}{\left({x}^{2}+1\right)}^{-\frac{3}{2}}\left(2x\right)\phantom{\rule{0ex}{0ex}}=-\frac{x}{{\left({x}^{2}+1\right)}^{\frac{3}{2}}}$