07Bdifftheory

= Theory of Differentiation =


 * 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.

math . \qquad \qquad \text{Gradient } = \dfrac {\Delta y}{\Delta x} = \dfrac {y_2 - y_1}{x_2 - x_1} \qquad. math


 * Note that D (delta) is the Greek symbol for capital D and is used to denote "change."

>>
 * ie D x is read as "change in x."
 * D x = x 2 – x 1




 * Using function notation, and using D x = h, the gradient rule becomes:

math . \qquad \qquad \dfrac {\Delta f}{\Delta x} = \dfrac {f(x+h) - f(x)}{h} \qquad. math


 * The ** instantaneous rate of change at a point **
 * otherwise known as the ** derivative **
 * is defined by the limit of this gradient as h → 0.


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

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


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


 * The alternative notation for derivative is:

math . \qquad \qquad \dfrac {dy}{dx} \text{ or sometimes } \dfrac {d}{dx}(y) \qquad. math


 * 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:

math . \qquad \dfrac{dy}{dx} \text{ is really just } \dfrac{rise}{run} \text{ only with very small steps in each direction} \qquad. math

** Example 1 **
... ... Using first principles, find the derivative of .. **//f(x) = x 2 + 2//**


 * Solution:**

math . \qquad f'(x) = \begin{matrix} \lim \\ h \to 0 \\ \end{matrix} \; \dfrac {f(x+h) - f(x)}{h} \qquad. math . math . \qquad \quad\quad = \begin{matrix} \lim \\ h \to 0 \\ \end{matrix} \; \dfrac { \big( (x+h)^2+2 \big) - \big( x^2+2 \big)}{h} \qquad. math . math . \qquad \quad\quad = \begin{matrix} \lim \\ h \to 0 \\ \end{matrix} \; \dfrac {x^2+2xh+h^2+2 - x^2-2}{h} \qquad. math . math . \qquad \quad\quad = \begin{matrix} \lim \\ h \to 0 \\ \end{matrix} \; \dfrac {2xh+h^2}{h} \qquad. math . math . \qquad \quad\quad = \begin{matrix} \lim \\ h \to 0 \\ \end{matrix} \; \dfrac {h \big( 2x+h \big)}{h} \qquad. math . math . \qquad \quad\quad = \begin{matrix} \lim \\ h \to 0\\ \end{matrix} \;\; 2x + h \qquad. math . math . \qquad \quad\quad= \;\; 2x \qquad. math

... ... The derivative of f(x) is therefore f '(x) = 2x.


 * 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 × x.

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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