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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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Consider :- \( \scriptsize y = \normalsize \frac{1}{(3x\; – \;4)^2 }\scriptsize = (3x -4)^{-2}\)

A function such as this sets a problem for differentiation.

It cannot be expressed as a polynomial. So we look at a very important technique devised to handle such functions.

To find \( \frac{dy}{dx} \) for the function above, we treat \( \scriptsize = (3x -4)^{-2} \! \! \! \) as a composite function built up in two stages from a core function (3x – 4), which we call u and then taking \( \scriptsize y = u^{-2} \) So we have u = (3x – 4)

Hence the general form of a function of function rule also called chain rule which is given as

\( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\)

Then in the above problem,

\( \scriptsize y = (3x -4)^{-2} \)

Solution

\( \scriptsize y = u^{-2}, u = (3x -4) \)

Thus \( \frac{dy}{du} = \scriptsize -2u^{-3} \normalsize = \frac{-2}{u^3} \)

while \( \frac{du}{dx} \scriptsize = 3\)

\( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\)

\( \frac{dy}{dx} = \; – \frac{2}{u^3} \times \scriptsize 3 =\; – \normalsize\frac{6}{u^3} \)

Substituting for u

= \(– \frac{6}{(3x \;-\; 4)^3} \)

Example:

2. Differentiate \( \scriptsize y = (4x -5)^6 \)

Let \( \scriptsize y = u^{6}, u = (4x – 5) \)

Thus \( \frac{dy}{du} = \scriptsize 6u^{5}\)

while \( \frac{du}{dx} \scriptsize = 4\)

\( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\)

= \(\scriptsize 6u^{5} \times 4 \)

\( \frac{dy}{dx} = \scriptsize 24u^{5} \)

and substituting u = 4x – 5 gives \( \frac{dy}{dx} = \scriptsize 24(4x-5)^{5} \)

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