### Differentiation: The derivative

### The difference quotient at a point

Using the difference quotient, we can approach the change at one point of a graph.

We want to approach the change at point #\green{x}=\green{2}# for the function #\blue{f(x)}=\blue{x^2}#. To do so, we take the difference quotient on an interval around #\green{2}#: \[[\green{2},\green{2}+\orange{h}],\]

in which we choose a smaller #\orange{h}# each time.

The smaller #\orange{h}# we choose, the better we can see the change at the point. We see that these values are getting closer and closer to #4# as we choose smaller values of #\orange{h}#. The change at a point is also called the **slope**.

We can approach the change at a point #x=a# for each function by determining the difference quotient on the interval #[a,a+h]#.

The difference quotient on an interval of length h

For a function #\blue{f}#, the **difference quotient** at point #\green{x}=\green{a}# with difference #\orange{h}# is defined as follows:

\[\dfrac{\Delta y}{\Delta x}= \dfrac{f(\green{a}+\orange{h})-f(\green{a})}{\orange{h}}\]

We can leave #\orange{h}# as is in our calculation.

**Example**

#\blue{f(x)}=\blue{x^2}# and #\green{a}=\green{4}# give:

\[\begin{array}{rcl}\dfrac{\Delta y}{\Delta x}&=& \dfrac{(\green{4}+\orange{h})^2-\green{4}^2}{\orange{h}}\\&=& \dfrac{16+8\cdot\orange{h}+\orange{h}^2-16}{\orange{h}} \\&=& \dfrac{8\cdot\orange{h}+\orange{h}^2 }{\orange{h}} \\&=& 8+\orange{h}\end{array}\]

#\begin{array}{rcl}\dfrac{\Delta y}{\Delta x}&=& \dfrac{f(5+h)-f(5)}{h}\\&&\phantom{xxx}\blue{\text{definition difference quotient at }x=5 }\\&=&\dfrac{2\cdot(5+h)+3-(2\cdot5+3)}{h}\\&&\phantom{xxx}\blue{5 \text{ entered in }f}\\

&=&\dfrac{10+2\cdot h -10}{h}\\&&\phantom{xxx}\blue{\text{expanded brackets}}\\&=&\dfrac{2\cdot h}{h}\\&&\phantom{xxx}\blue{\text{added}}\\&=&2\\&&\phantom{xxx}\blue{\text{eliminated }h}\end{array}#

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