Log In Start studying!

Select your language

Suggested languages for you:

Americas

English (US)

Europe

Q. 90

Expert-verified

Calculus

Found in: Page 200

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

Answers without the blur.

Just sign up for free and you're in.

Register for Free I’ll do it later

Short Answer

The following reciprocal rules tells us hoe to differentiate the reciprocal of a function
$\frac{d}{d x} (\frac{1}{f (x)}) = - \frac{1}{{[f (x)]}^{2}}$
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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given information

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

Step 2: Prove using the definition

We have

$\lim_{h \to 0} \frac{\frac{1}{f (x + h)} - \frac{1}{f (x)}}{h} \lim_{h \to 0} \frac{f (x) - f (x + h)}{f (x) f (x + h) h} - \lim_{h \to 0} \frac{f (x + h) - f (x)}{f (x) (f (x + h)) h} = - \frac{f' (x)}{{[f (x)]}^{2}}$

Step 3: Prove using product rule

We have

$\frac{d}{d x} [f (x) g (x)] = f (x) g' (x) + f' (x) g (x) \frac{d}{d x} [\frac{1}{f (x)} \cdot 1] = \frac{1}{f (x)} \cdot (0) + [- \frac{f' (x)}{f {(x)}^{2}}] \frac{d}{d x} [\frac{1}{f (x)}] = - \frac{f' (x)}{f {(x)}^{2}}$

Most popular questions for Math Textbooks

Differentiate $f (x) = {(3 x + \sqrt{x})}^{2}$ in three ways. When you have completed all three parts, show that your three answers are the same:

(a) with the chain rule

(b) with the product rule but not the chain rule

(c) without the chain or product rules.

A bowling ball dropped from a height of $400$ feet will be $s (t) = 400 - 16 t^{2}$ feet from the ground after $t$ seconds. Use a sequence of average velocities to estimate the instantaneous velocities described below:

When the bowling ball is first dropped, with $h = 0.5,$ $h = 0.25, a n d h = 0.1$

Use the definition of the derivative to find the equations of the lines described in Exercises 59-64.

The line tangent to the graph of $y = 4 x + 3$ at the point $(- 2, - 5)$

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.

$f (x) = \sqrt{3 x - 4 {(2 x + 1)}^{6}}$

Each graph in Exercises 31–34 can be thought of as the associated slope function f' for some unknown function f. In each case sketch a possible graph of f.

Want to see more solutions like these?

Sign up for free to discover our expert answers

Get Started - It’s free

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free