Introduction to differentiation: Definition of differentiation
The notion of derivative
In the previous section, we defined the notion of difference quotient. We approximated the slope of the tangent line to a graph by taking the difference quotient of the function at the point #a# with a small difference #h#. In this section we will see what happens when we make #h# increasingly smaller. We will discover that the slope of the tangent line to a graph at a point is equal to the limit for #h# to #0# of the difference quotient.
But first we will see what happens when #h# becomes smaller and smaller.
Approximate the slope of #l# at the point #\rv{1,3.12500}# by calculating the difference quotient of #f# at #1# with difference #h# for #h=1#, #h=\frac{1}{10}#, #h=\frac{1}{100}#, #h=\frac{1}{1000}#, and #h=\frac{1}{10000}#, respectively. Give your answer to #5# decimal places.
The difference quotient for #h=\frac{1}{10}# is: #0.26250#
The difference quotient for #h=\frac{1}{100}# is: #0.25125#
The difference quotient for #h=\frac{1}{1000}# is: #0.25012#
The difference quotient for #h=\frac{1}{10000}# is: #0.25001#
This follows from the following calculations:
\[\begin{array}{r|cl}
h&&\text{difference quotient of }f\text{ at }1\text{ with difference }h\\
\hline
1&\phantom{xx}& \frac{f(2)-3.12500}{1}\phantom{0000}=\frac{3.50000-3.12500}{1}=0.37500\\
\frac{1}{10}&&\frac{f(1.10000)-3.12500}{\frac{1}{10}}=\frac{3.15125-3.12500}{\frac{1}{10}}=0.26250\\
\frac{1}{100}&&\frac{f(1.01000)-3.12500}{\frac{1}{100}}=\frac{3.12751-3.12500}{\frac{1}{100}}=0.25125\\
\frac{1}{1000}&&\frac{f(1.00100)-3.12500}{\frac{1}{1000}}=\frac{3.12525-3.12500}{\frac{1}{10}}=0.25012\\
\frac{1}{10000}&&\frac{f(1.00010)-3.12500}{\frac{1}{10000}}=\frac{3.12503-3.12500}{\frac{1}{10000}}=0.25001
\end{array}\]
We see that as #h# becomes smaller and smaller, the difference quotient approximates the slope of the tangent line more and more accurately. This leads to the following definition of the slope of a graph, which we call derivative.
Differentiation
Let #f# be a function defined on an interval around a point #a#. If #\lim_{h\to 0}\frac{f(a+h)−f(a)}{h}# exists, then #f# is called differentiable at #a#; the limit is called the derivative of #f# at #a#. This limit is indicated by #f’(a)# and also by #\dfrac{{\dd f}}{{\dd}x}(a)#. The act of determining the derivative is called differentiation.
If #f# is differentiable at all points of an interval #I#, we say that #f# differentiable on #I#. In that case, #f’# is a function on #I#.
The value #f'(x)# is often indicated by #\frac{{\dd }}{{\dd}x}f(x)#.
If #y# is a function rule of #f#, we also write #\left.\frac{{\dd }}{{\dd}x}y\right|_{x=a}# instead of #f'(a)# or #\frac{{\dd f }}{{\dd}x}(a)#.
The number #f'(a)# is the slope of the graph of #f# at the point #\rv{a,f(a)}#.
Often the function #f# and the function rule #f(x)# (which is the value of #f# at any point #x#) are used interchangeably. The expressions #f'(x)# and #\dfrac{\dd}{\dd x}f(x)# are also used instead of #f'#.
An example of the use of the vertical bar with #f(x)=x^2+1# is\[\frac{\dd f}{\dd x}(3)=\left.\frac{\dd}{\dd x}(x^2+1)\right|_{x=3}=\left.(2x)\right|_{x=3}=6\tiny.\]
Be aware that in general not every function is differentiable; in this course, this will not be discussed in greater detail.
Using this definition, we can calculate the derivative at a point. This is done in two steps. In the first step, we write down the difference quotient at the point with difference #h#. In the second step, we let #h# go to #0#, that is: we take the limit #h\to 0#. The examples below show these two steps, first at a specific point, then at a general point #x=a#.
First, we calculate the difference quotient of #f# at #1# with difference #h#:
\[\begin{array}{rcl}\frac{\Delta y}{\Delta x}&=&\frac{f(1+h)-f(1)}{h}\\
&&\phantom{xx}\color{blue}{\text{definition}}\\
&=&\frac{\frac{1}{5} \cdot (1+h)^2+6-(\frac{1}{5} \cdot (1)^2+6)}{h}\\
&&\phantom{xx}\color{blue}{f(x)=\frac{1}{5}x^2+6}\\
&=&\frac{\frac{1}{5} \cdot (1^2+h^2+2 \cdot h)-\frac{1}{5} \cdot (1)^2}{h}\\
&&\phantom{xx}\color{blue}{\text{expanded}}\\&=&\frac{\frac{1}{5} \cdot (h^2+2 \cdot h)}{h}={{h}\over{5}}+{{2}\over{5}}\\&&\phantom{xx}\color{blue}{\text{simplified}}\\\end{array}\]
Next, we calculate the derivative of #f# at the point #x=1# as the limit for #h \to 0# of the difference quotient:\[f'(1)= \lim_{h \to 0}\left ({{h}\over{5}}+{{2}\over{5}}\right)={{2}\over{5}}\tiny.\]
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