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Q. 90

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Found in: Page 200

### Calculus

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

# The following reciprocal rules tells us hoe to differentiate the reciprocal of a function$\frac{d}{dx}\left(\frac{1}{f\left(x\right)}\right)=-\frac{1}{{\left[f\left(x\right)\right]}^{2}}$Prove this usinga) definition of the derivativeb) by using the quotient rule

We prove the reciprocal rule using definition of derivative and quotient rule

See the step by step solution

## Step 1: Given information

We are given the reciprocal rule as $\frac{d}{dx}\left(\frac{1}{f\left(x\right)}\right)=-\frac{1}{{\left[f\left(x\right)\right]}^{2}}$

We have

## Step 3: Prove using product rule

We have

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