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SS3: MATHEMATICS - 2ND TERM

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  1. Matrices I | Week 1
    6 Topics
  2. Matrices II | Week 2
    1 Topic
    |
    1 Quiz
  3. Commercial Arithmetic | Week 3
    7 Topics
    |
    1 Quiz
  4. Coordinate Geometry | Week 4
    8 Topics
    |
    1 Quiz
  5. Differentiation of Algebraic Expressions | Week 5 & 6
    7 Topics
  6. Application of Differentiation | Week 7
    4 Topics
    |
    1 Quiz
  7. Integration | Week 8
    8 Topics
    |
    1 Quiz



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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, \( \frac{dv}{dx} = \scriptsize 3\)

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

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

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

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

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