Here we study the derivative of a function, as a function, in its own right.
1 The derivative of a function, as a function
First, we have to find an alternate definition for \(f'(a)\), the derivative of a function \(f\) at
\(a\).
Let’s start with the average rate of change of the function \(f\) as the input
changes from \(a\) to \(x\). We will introduce a new variable, \(h\), to denote the
difference between \(x\) and \(a\). That is \(x-a=h\) or \(x=a+h\). Take a look at the figure below.
Now
we can write
\[ {\text {average rate of change }}=\frac {f(a+h)-f(a)}{(a+h)-a}=\frac {f(a+h)-f(a)}{h} \]
What happens if \(h\to 0\)? In other words, what is the meaning of the limit
\[ \lim _{h\to 0} \frac {f(a+h)-f(a)}{h}?\]
This limit represents \(f'(a)\), the instantaneous rate of change of \(f\) at \(a\)! Therefore, we have an
alternate way of writing the definition of the derivative at the point \(a\), namely
\[ f'(a) = \lim _{h\to 0} \frac {f(a+h)-f(a)}{h}. \]
Let
\(f(x) = x^2-2x\). Using the alternate expression for the derivative, find the slope of the tangent
line to the curve
\(y=f(x)\) at the point
\((2,f(2))\).
The slope of the tangent line is given by the
derivative,
\(f'(2)\).
\[ f'(2) = \lim _{h\to 0}\frac {f\left (\answer [given]{2+h}\right )-f\left (2\right )}{h}. \]
Now substitute in for the function we know,
\[ f'(2) = \lim _{h\to 0}\frac {(2+h)^2-2(2+h) -0}{h}.\]
Now expand the
numerator of the fraction,
\[ f'(2) =\lim _{h\to 0} \frac {4+4h+h^2-4-2h }{h}. \]
Now combine like-terms,
\[ f'(2) = \lim _{h\to 0} \frac {2h+h^2}{h}.\]
Factor an
\(h\) from every term in
the numerator,
\[ f'(2) = \lim _{h\to 0}\frac {\cancel {h}\left (2+h\right )}{\cancel {h}}. \]
Compute the limit,
\[ f'(2) = \lim _{h\to 0}(2+h)=\answer [given]{2}. \]
This alternate definition of the derivative of \(f\) at \(a\), namely,
\[ f'(a) = \lim _{h\to 0}\frac {f(a+h)-f(a)}{h}, \]
(provided that the
limit exists), allows us to define \(f'(x)\) for any value of \(x\),
\[ f'(x) = \lim _{h\to 0}\frac {f(x+h)-f(x)}{h},\]
(provided that the limit
exists).
This is how we define a new function, \(f'\), the derivative of \(f\). The domain of \(f'\) consists of
all points in the domain of \(f\) where the function \(f\) is differentiable. The value \(f'(x)\)
gives us the instantaneous rate of change of \(f\) at any point \(x\) in the domain of
\(f'\).
Given a function \(f\) from some set of real numbers to the real numbers, the derivative \(f'\)
is also a function from some set of real numbers to the real numbers. Understanding
the relationship between the functions \(f\) and \(f'\) helps us understand any situation (real
or imagined) involving changing values.
Given the function
\(f(x) = 3x+2\), find
\(f'(x)\).
Start with the definition of
\(f'(x)\) \[ f'(x) = \lim _{h\to 0}\frac {f(x+h)-f(\answer [given]{x})}{h}. \]
Replace
\(f\) with its formula,
\[ f'(x) = \lim _{h\to 0}\frac {3(x+h)+2-\left (\answer [given]{3x+2}\right )}{h}. \]
Simplify the numerator,
\[ f'(x) = \lim _{h\to 0}\frac {3\cancel {h}}{\cancel {h}}. \]
Evaluate the limit.
\[ f'(x) = \lim _{h\to 0}{3}=\answer [given]{3}. \]
Given the function
\(f(x) = |x|\), find
\(f'(x)\).
Recall, the domain of
\(f\) is
\((-\infty , \infty )\) and
\(f\) can be considered a piecewise defined function, since
\[ f(x) = |x|= \begin{cases} \answer [given]{-x} &\text {if $x<0$,}\\ \answer [given]{x} &\text {if $x\ge 0$}. \end{cases} \]
We will first compute
\(f'(x)\) when
\(x>0\). Start with the definition of
\(f'(x)\) \[ f'(x) = \lim _{h\to 0}\frac {f(x+h)-f(\answer [given]{x})}{h}.\]
Replace
\(f\) with its formula,
\[ f'(x) = \lim _{h\to 0}\frac {|x+h|-\answer [given]{|x|}}{h}. \]
\[ f'(x) = \lim _{h\to 0}\frac {x+h-\answer [given]{x}}{h}.\]
Note: When
\(x>0\), then for all small enough values of
\(h\) it follows that
\(x+h>0\). Therefore,
\(|x+h|=x+h\). Now
we have
\[ f'(x) = \lim _{h\to 0}\frac {h}{h}= \lim _{h\to 0}\answer [given]{1}=\answer [given]{1}.\]
Next we compute the derivative \(f'(x)\) when \(x<0\).
\[ f'(x) = \lim _{h\to 0}\frac {f(x+h)-f(\answer [given]{x})}{h}. \]
Replace
\(f\) with its formula,
\[ f'(x) = \lim _{h\to 0}\frac {|x+h|-\answer [given]{|x|}}{h}. \]
\[ f'(x) = \lim _{h\to 0}\frac {-x-h-\answer [given]{-x}}{h}. \]
Note: When
\(x<0\),
then for all small enough values of
\(h\) it follows that
\(x+h<0\). Therefore,
\(|x+h|=-(x+h)=-x-h\). That gives
\[ f'(x) = \lim _{h\to 0}\frac {-h}{h}= \lim _{h\to 0}\answer [given]{-1}=\answer [given]{-1}.\]
What remains to be done is to check whether the derivative \(f'(0)\) exists. Start with the
definition of \(f'(0)\)
\[ f'(0) = \lim _{h\to 0}\frac {f(0+h)-f(\answer [given]{0})}{h}. \]
Replace
\(f\) with its formula,
\[ f'(0) = \lim _{h\to 0}\frac {|0+h|-\answer [given]{|0|}}{h}.\]
\[ f'(0) = \lim _{h\to 0}\frac {|h|}{h}. \]
Note: When
\(h>0\), then
\(|h|=h\), but when
\(h<0\), then
\(|h|=-h\).
Therefore, instead of computing the limit above, we will compute the two one-sided
limits and compare them.
\[ \lim _{h\to 0^+}\frac {|h|}{h}= \lim _{h\to 0^+}\frac {h}{h}=\lim _{h\to 0^+}\answer [given]{1}=\answer [given]{1}; \]
\[ \lim _{h\to 0^-}\frac {|h|}{h}= \lim _{h\to 0^-}\frac {-h}{h}=\lim _{h\to 0^-}\answer [given]{-1}=\answer [given]{-1};\]
Since the two one-sided limits are not equal it follows that
\[ \lim _{h\to 0}\frac {|h|}{h} {\text { DOES NOT EXIST!}}\]
Therefore,
\(f'(0)\)
DOES NOT EXIST, which means that
\(f\) is NOT DIFFERENTIABLE at
\(x=0\)!
To summarize
\[ f'(x) = \begin{cases} \answer [given]{-1} &\text {if $x<0$,}\\ \answer [given]{1} &\text {if $x> 0$}. \end{cases} \]
Is it true that for any function
\(f\) the domain of
\(f'\) is equal to the domain of
\(f\)?
yes no
This is not true. Consider the function
\(f(x)=|x|\). The domain of
\(f\) is
\((-\infty , \infty )\) and the domain of
\(f'\) is
\((-\infty ,0)\cup (0,\infty )\)
This example demonstrates that a function \(f\) and its derivative, \(f'\), may have different
domains.
Can two different functions, say,
\(f\) and
\(g\), have the same derivative?
yes no
Many different functions can share the same derivatives. Consider two different
functions, \(f\) and \(g\), defined by
\(f(x)=x\) and \(g(x)=x+5\). Then, \(f'(x)=1\), and \(g'(x)=1\), for all real numbers \(x\).
So, the derivatives of these two different functions are equal.
Let’s compare the graphs of \(f\) and \(f'\) for the derivatives we’ve computed so
far:
\[f(x)=x^2, f'(x)=2x\]
\[f(x)=3x+2, f'(x)=3\]
\[f(x)=|x|, f'(x)=\begin{cases} 1 & \text {for } x>0 \\ -1 & \text {for } x<0 \end{cases}\]
For each of the three pairs of functions, describe
\(y=f(x)\) when
\(f'\) is positive, and when
\(f'\) is
negative.
When
\(f'\) is positive,
\(y=f(x)\) is
positiveincreasingnegativedecreasing. When
\(f'\) is
negative,
\(y=f(x)\) is
positiveincreasingnegativedecreasing Here we see the graph of
\(f'\), the derivative of some function
\(f\).
Which of the following graphs could be \(y = f(x)\)?
![[Picture]](digInTheDerivativeAsAFunction.online-b154a5eebbc045948159911e0918b620.svg)
![[Picture]](digInTheDerivativeAsAFunction.online-e18e58dd84c53c434cef8a085e6e41b2.svg)
![[Picture]](digInTheDerivativeAsAFunction.online-c82e78bb2ed11927397293547444c1fd.svg)
![[Picture]](digInTheDerivativeAsAFunction.online-2fab1fc2a1e219e4cdaf8cec7d287284.svg)
![[Picture]](digInTheDerivativeAsAFunction.online-08041e52832206fea11ef94232cd5568.svg)
![[Picture]](digInTheDerivativeAsAFunction.online-7b4d8802b85175fd476b4f42ce26546f.svg)
![[Picture]](digInTheDerivativeAsAFunction.online-7b4dd19a0d2d788d2436758e52dccddd.svg)
![[Picture]](digInTheDerivativeAsAFunction.online-3932ed7b3765e3ceee7ce9c351356e36.svg)