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, \( \frac{dv}{dx} = \scriptsize 3\)
\( \frac{dy}{dx} = \frac {v \frac{du}{dx} \; – \; u \frac{dv}{dx}}{v^2} \) \( \frac{dy}{dx} = \frac{ (3x-2)2 \; – \;( 2x + 1)3}{(3x – 2)^2}\) \( \frac{dy}{dx} = \frac{ (6x-4) \; – \;(6x + 3)} {(3x – 2)^2}\) \( \frac{dy}{dx} = \frac{-7} {(3x – 2)^2}\)
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