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Lesson 5, Topic 7
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# Quotient Rule

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In the case of quotient, if u and v are functions of x and $$\scriptsize y = \normalsize \frac{u}{v}$$

then $$\frac{dy}{dx} = \frac {v \frac{du}{dx} \; – \; u \frac{dv}{dx}}{v^2}$$

Note: Do not use the quotient rule when the denominator is a single term

E.g $$\frac{3x^2\; – x}{2x^2}$$

rewrite it as $$\frac{3}{2} – \frac{1}{2x}$$ and differentiate

Example:

Differentiate $$\scriptsize y = \normalsize \frac{2x + 1 }{3x – 2}$$ with respect to x

Solution:

Let $$\scriptsize y = \normalsize \frac{u}{v}$$

Therefore,

u = 2x + 1, $$\frac{du}{dx} = \scriptsize 2$$

v = 3x – 2, $$\frac{dv}{dx} = \scriptsize 3$$

$$\frac{dy}{dx} = \frac {v \frac{du}{dx} \; – \; u \frac{dv}{dx}}{v^2}$$

$$\frac{dy}{dx} = \frac{ (3x-2)2 \; – \;( 2x + 1)3}{(3x – 2)^2}$$

$$\frac{dy}{dx} = \frac{ (6x-4) \; – \;(6x + 3)} {(3x – 2)^2}$$

$$\frac{dy}{dx} = \frac{-7} {(3x – 2)^2}$$

error: