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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 general if, y =f(x) then f'(x) = \( \frac{dy}{dx} \) is the limiting value of \( \frac{f(x + \delta x) – f(x)}{\delta x} \)

where \( \scriptsize \delta x \rightarrow 0 \)

We assume that this limiting value will exist and can be found.

We write\( \frac{dy}{dx} = \scriptsize \lim\limits_{\delta x \to 0} \normalsize \frac{f(x + \delta x) – f(x)}{\delta x} \)

where \(\scriptsize \lim\limits_{\delta x \to 0} \) means the limiting value as \( \scriptsize \delta x \rightarrow 0 \)

Example:

Differentiate \( \scriptsize 6x^2 – 1 \) with respect to x from first principles.

We write \( \scriptsize 6x^2 – 1 \)

Working from first principle, any increase in x to x + δx produces a corresponding increase in y to y + δy

Then \( \scriptsize y + \delta y = 6(x + \delta x)^2 – 1 \)

\( \scriptsize \delta y = 6(x + \delta x)^2 – 1 – y \)

\( \scriptsize \delta y = 6[x^2 + 2x \delta x + (\delta x)^2] – 1 – (6x^2 – 1) \)

\( \scriptsize \delta y = 6x^2 + 12x \delta x + 6(\delta x)^2 – 1 – 6x^2 + 1 \)

\( \scriptsize \delta y = 12x \delta x + 6(\delta x)^2 \)

Divide through by δx, then

\( \frac{ \delta y }{\delta x} =\scriptsize 12x + 6(\delta x) \)

Now consider \( \scriptsize \delta x \rightarrow 0 \) and so

\( \frac{dy }{dx} = \scriptsize \lim\limits_{\delta x \to 0} \normalsize\frac{ \delta y }{\delta x} = \scriptsize 12x + 6(\delta x) \)

\( \frac{dy }{dx} = \scriptsize 12x \)

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