Suggested languages for you:

Americas

Europe

Q. 90

Expert-verifiedFound in: Page 200

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

$\frac{d}{dx}\left(\frac{1}{f\left(x\right)}\right)=-\frac{1}{{\left[f\right(x\left)\right]}^{2}}$

Prove this using

a) definition of the derivative

b) by using the quotient rule

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

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

We have

$\underset{h\to 0}{\mathrm{lim}}\frac{\frac{1}{f(x+h)}-\frac{1}{f\left(x\right)}}{h}\phantom{\rule{0ex}{0ex}}\underset{h\to 0}{\mathrm{lim}}\frac{f\left(x\right)-f(x+h)}{f\left(x\right)f(x+h)h}\phantom{\rule{0ex}{0ex}}-\underset{h\to 0}{\mathrm{lim}}\frac{f(x+h)-f\left(x\right)}{f\left(x\right)\left(f\right(x+h\left)\right)h}\phantom{\rule{0ex}{0ex}}=-\frac{f\text{'}\left(x\right)}{{\left[f\right(x\left)\right]}^{2}}$

We have

$\frac{d}{dx}\left[f\right(x\left)g\right(x\left)\right]=f\left(x\right)g\text{'}\left(x\right)+f\text{'}\left(x\right)g\left(x\right)\phantom{\rule{0ex}{0ex}}\frac{d}{dx}[\frac{1}{f\left(x\right)}\xb71]=\frac{1}{f\left(x\right)}\xb7\left(0\right)+[-\frac{f\text{'}\left(x\right)}{f{\left(x\right)}^{2}}]\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left[\frac{1}{f\left(x\right)}\right]=-\frac{f\text{'}\left(x\right)}{f{\left(x\right)}^{2}}$

94% of StudySmarter users get better grades.

Sign up for free