05 4firstprins

# Theory of Differentiation

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For any function, the average rate of change between two points is the same as the gradient of a straight line between those two points.

… … Gradient $= \dfrac {\Delta y}{\Delta x} = \dfrac {y_2 - y_1}{x_2 - x_1}$

Note $\Delta$ (delta) is the Greek symbol for capital D and is used to denote "change."

… … ie $\Delta x$ is read as "change in x."
… … $\Delta x = x_2 - x_1$

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Using function notation, and using $\Delta x = h$, the gradient rule becomes:

… … $\dfrac {\Delta f}{\Delta x} = \dfrac {f(x+h) - f(x)}{h}$

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• The instantaneous rate of change at a point
• otherwise known as the derivative
• is defined by the limit of the above gradient rule as $h \rightarrow 0$.

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… … $f'(x) = \begin{matrix} \lim \\ h \to 0 \\ \end{matrix} \; \dfrac {f(x+h) - f(x)}{h}$

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## Derivative by First Principles

• The instantaneous rate of change at a point
• also represents the gradient of the tangent to the curve at that point.

… … $f'(x) = \begin{matrix} \lim \\ h \to 0 \\ \end{matrix} \; \dfrac {f(x+h) - f(x)}{h}$

This gradient function is usually referred to as the derivative by first principles.

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The alternative notation for derivative is: … $\dfrac {dy}{dx} \text{ or sometimes } \dfrac {d}{dx}(y)$

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In the derivative,

• dx can be read as "a really small change in x" … and
• dy can be read as "a really small change in y",

so:
… … $\dfrac{dy}{dx} \text{ is really just } \dfrac{rise}{run}$ only with very small steps in each direction

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#### Example 1

… … Using first principles, find the derivative of .. $f(x) = x^2 + 2$

Solution:

… … $f'(x) = \begin{matrix} \lim \\ h \to 0 \\ \end{matrix} \; \dfrac {f(x+h) - f(x)}{h}$

… … … .. $= \begin{matrix} \lim \\ h \to 0 \\ \end{matrix} \; \dfrac { \big( (x+h)^2+2 \big) - \big( x^2+2 \big)}{h}$

… … … … $= \begin{matrix} \lim \\ h \to 0 \\ \end{matrix} \; \dfrac {x^2+2xh+h^2+2 - x^2-2}{h}$

… … … … $= \begin{matrix} \lim \\ h \to 0 \\ \end{matrix} \; \dfrac {2xh+h^2}{h}$

… … … … $= \begin{matrix} \lim \\ h \to 0 \\ \end{matrix} \; \dfrac {h \big( 2x+h \big)}{h}$

… … … … $= \begin{matrix} \lim \\ h \to 0\\ \end{matrix} \;\; 2x + h$

… … … … $= \;\; 2x$
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… … … … The derivative of $f(x) = x^2 + 2$ is therefore $f'(x) = 2x$.

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• This means that the instantaneous rate of change
• ie the gradient of the tangent
• of any point on the graph of f(x) can be found using the rule: $f'(x) = 2 \times x$.

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Note:

A function can only be differentiated at a point if its graph is continuous and smooth at that point.

• Also, we can not differentiate the endpoint of a function.
• This is because differentiating is equivalent to finding a tangent to the curve at that point
• If the graph is discontinuous, or an endpoint, or not smooth then the tangent is undefined.

See here for the conditions of differentiation.

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