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Lesson 5, Topic 2
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# Differentiation from First Principle

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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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