Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In Exercise 89of the Section 2.3you used the definition of derivative to prove the quotient rule. Prove it now another way: by writing a quotient fg as a product and applying the product, power, and chain rules. Point out where you use each rule.

Short Answer

Expert verified

y=fgyg=fddx(yg)=ddx(f)ydgdx+gdydx=dfdxgdydx=dfdx-ydgdxdydx=dfdx-ydgdxgddxfg=dfdx-fgdgdxgddxfg=gdfdx-fdgdxg2

Hence proved.

Step by step solution

01

Step 1. Given Information

We have to prove the quotient rule of derivative.

That is we have to prove that :-

ddxfg=gdfdx-fdgdxg2

We have to convert quotient to a product. Then we have to use product, quotient and chain rule to prove this rule.

02

Step 2. Proof of quotient rule

Consider the following quotient :-

y=fgyg=f

Take derivative on both sides, then we have :-

ddx(yg)=ddx(f)

Now use product and chain rule, then we have :-

ydgdx+gdydx=dfdxgdydx=dfdx-ydgdxdydx=dfdx-ydgdxg

Put the value of y=fg, then we have :-

role="math" localid="1648663599119" ddxfg=dfdx-fgdgdxgddxfg=gdfdx-fdgdxg2

This is the required result.

Hence the quotient rule is proved.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free