### 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{1,3.50000}# 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: #1.05000#

The difference quotient for #h=\frac{1}{100}# is: #1.00500#

The difference quotient for #h=\frac{1}{1000}# is: #1.00050#

The difference quotient for #h=\frac{1}{10000}# is: #1.00005#

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)-3.50000}{1}\phantom{0000}=\frac{5.00000-3.50000}{1}=1.50000\\

\frac{1}{10}&&\frac{f(1.10000)-3.50000}{\frac{1}{10}}=\frac{3.60500-3.50000}{\frac{1}{10}}=1.05000\\

\frac{1}{100}&&\frac{f(1.01000)-3.50000}{\frac{1}{100}}=\frac{3.51005-3.50000}{\frac{1}{100}}=1.00500\\

\frac{1}{1000}&&\frac{f(1.00100)-3.50000}{\frac{1}{1000}}=\frac{3.50100-3.50000}{\frac{1}{10}}=1.00050\\

\frac{1}{10000}&&\frac{f(1.00010)-3.50000}{\frac{1}{10000}}=\frac{3.50010-3.50000}{\frac{1}{10000}}=1.00005

\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 #3# with difference #h#:

\[\begin{array}{rcl}\frac{\Delta y}{\Delta x}&=&\frac{f(3+h)-f(3)}{h}\\

&&\phantom{xx}\color{blue}{\text{definition}}\\

&=&\frac{\frac{1}{5} \cdot (3+h)^2+6-(\frac{1}{5} \cdot (3)^2+6)}{h}\\

&&\phantom{xx}\color{blue}{f(x)=\frac{1}{5}x^2+6}\\

&=&\frac{\frac{1}{5} \cdot (3^2+h^2+2 \cdot 3h)-\frac{1}{5} \cdot (3)^2}{h}\\

&&\phantom{xx}\color{blue}{\text{expanded}}\\&=&\frac{\frac{1}{5} \cdot (h^2+2 \cdot 3h)}{h}={{h}\over{5}}+{{6}\over{5}}\\&&\phantom{xx}\color{blue}{\text{simplified}}\\\end{array}\]

Next, we calculate the derivative of #f# at the point #x=3# as the limit for #h \to 0# of the difference quotient:\[f'(3)= \lim_{h \to 0}\left ({{h}\over{5}}+{{6}\over{5}}\right)={{6}\over{5}}\tiny.\]

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