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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.

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