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(6+h)-f(6)}{h}\\&&\phantom{xxx}\blue{\text{definition difference quotient at }x=6 }\\&=&\dfrac{5\cdot(6+h)+3-(5\cdot6+3)}{h}\\&&\phantom{xxx}\blue{6 \text{ entered in }f}\\
&=&\dfrac{30+5\cdot h -30}{h}\\&&\phantom{xxx}\blue{\text{expanded brackets}}\\&=&\dfrac{5\cdot h}{h}\\&&\phantom{xxx}\blue{\text{added}}\\&=&5\\&&\phantom{xxx}\blue{\text{eliminated }h}\end{array}#
Or visit omptest.org if jou are taking an OMPT exam.
