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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