# 3.2 The derivative as a function

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• Define the derivative function of a given function.
• Graph a derivative function from the graph of a given function.
• State the connection between derivatives and continuity.
• Describe three conditions for when a function does not have a derivative.
• Explain the meaning of a higher-order derivative.

As we have seen, the derivative of a function at a given point gives us the rate of change or slope of the tangent line to the function at that point. If we differentiate a position function at a given time, we obtain the velocity at that time. It seems reasonable to conclude that knowing the derivative of the function at every point would produce valuable information about the behavior of the function. However, the process of finding the derivative at even a handful of values using the techniques of the preceding section would quickly become quite tedious. In this section we define the derivative function and learn a process for finding it.

## Derivative functions

The derivative function gives the derivative of a function at each point in the domain of the original function for which the derivative is defined. We can formally define a derivative function as follows.

## Definition

Let $f$ be a function. The derivative function    , denoted by ${f}^{\prime },$ is the function whose domain consists of those values of $x$ such that the following limit exists:

${f}^{\prime }\left(x\right)=\underset{h\to 0}{\text{lim}}\frac{f\left(x+h\right)-f\left(x\right)}{h}.$

A function $f\left(x\right)$ is said to be differentiable at $a$    if $f\left(a\right)$ exists. More generally, a function is said to be differentiable on $S$    if it is differentiable at every point in an open set $S,$ and a differentiable function    is one in which ${f}^{\prime }\left(x\right)$ exists on its domain.

In the next few examples we use [link] to find the derivative of a function.

## Finding the derivative of a square-root function

Find the derivative of $f\left(x\right)=\sqrt{x}.$

Start directly with the definition of the derivative function. Use [link] .

$\begin{array}{ccccc}\hfill {f}^{\prime }\left(x\right)& =\underset{h\to 0}{\text{lim}}\frac{\sqrt{x+h}-\sqrt{x}}{h}\hfill & & & \begin{array}{c}\text{Substitute}\phantom{\rule{0.2em}{0ex}}f\left(x+h\right)=\sqrt{x+h}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}f\left(x\right)=\sqrt{x}\hfill \\ \text{into}\phantom{\rule{0.2em}{0ex}}{f}^{\prime }\left(x\right)=\underset{h\to 0}{\text{lim}}\frac{f\left(x+h\right)-f\left(x\right)}{h}.\hfill \end{array}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{\sqrt{x+h}-\sqrt{x}}{h}·\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\hfill & & & \begin{array}{c}\text{Multiply numerator and denominator by}\hfill \\ \sqrt{x+h}+\sqrt{x}\phantom{\rule{0.2em}{0ex}}\text{without distributing in the}\hfill \\ \text{denominator.}\hfill \end{array}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{h}{h\left(\sqrt{x+h}+\sqrt{x}\right)}\hfill & & & \text{Multiply the numerators and simplify.}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{1}{\left(\sqrt{x+h}+\sqrt{x}\right)}\hfill & & & \text{Cancel the}\phantom{\rule{0.2em}{0ex}}h.\hfill \\ & =\frac{1}{2\sqrt{x}}\hfill & & & \text{Evaluate the limit.}\hfill \end{array}$

## Finding the derivative of a quadratic function

Find the derivative of the function $f\left(x\right)={x}^{2}-2x.$

Follow the same procedure here, but without having to multiply by the conjugate.

$\begin{array}{ccccc}\hfill {f}^{\prime }\left(x\right)& =\underset{h\to 0}{\text{lim}}\frac{\left({\left(x+h\right)}^{2}-2\left(x+h\right)\right)-\left({x}^{2}-2x\right)}{h}\hfill & & & \begin{array}{c}\text{Substitute}\phantom{\rule{0.2em}{0ex}}f\left(x+h\right)={\left(x+h\right)}^{2}-2\left(x+h\right)\phantom{\rule{0.2em}{0ex}}\text{and}\hfill \\ f\left(x\right)={x}^{2}-2x\phantom{\rule{0.2em}{0ex}}\text{into}\hfill \\ {f}^{\prime }\left(x\right)=\underset{h\to 0}{\text{lim}}\frac{f\left(x+h\right)-f\left(x\right)}{h}.\hfill \end{array}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{{x}^{2}+2xh+{h}^{2}-2x-2h-{x}^{2}+2x}{h}\hfill & & & \text{Expand}\phantom{\rule{0.2em}{0ex}}{\left(x+h\right)}^{2}-2\left(x+h\right).\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{2xh-2h+{h}^{2}}{h}\hfill & & & \text{Simplify.}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{h\left(2x-2+h\right)}{h}\hfill & & & \text{Factor out}\phantom{\rule{0.2em}{0ex}}h\phantom{\rule{0.2em}{0ex}}\text{from the numerator.}\hfill \\ & =\underset{h\to 0}{\text{lim}}\left(2x-2+h\right)\hfill & & & \text{Cancel the common factor of}\phantom{\rule{0.2em}{0ex}}h.\hfill \\ & =2x-2\hfill & & & \text{Evaluate the limit.}\hfill \end{array}$

Find the derivative of $f\left(x\right)={x}^{2}.$

${f}^{\prime }\left(x\right)=2x$

We use a variety of different notations to express the derivative of a function. In [link] we showed that if $f\left(x\right)={x}^{2}-2x,$ then ${f}^{\prime }\left(x\right)=2x-2.$ If we had expressed this function in the form $y={x}^{2}-2x,$ we could have expressed the derivative as ${y}^{\prime }=2x-2$ or $\frac{dy}{dx}=2x-2.$ We could have conveyed the same information by writing $\frac{d}{dx}\left({x}^{2}-2x\right)=2x-2.$ Thus, for the function $y=f\left(x\right),$ each of the following notations represents the derivative of $f\left(x\right)\text{:}$

${f}^{\prime }\left(x\right),\text{}\phantom{\rule{0.2em}{0ex}}\frac{dy}{dx},\text{}\phantom{\rule{0.2em}{0ex}}{y}^{\prime },\text{}\phantom{\rule{0.2em}{0ex}}\frac{d}{dx}\left(f\left(x\right)\right).$

why n does not equal -1
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
proof the formula integration of udv=uv-integration of vdu.?
Find derivative (2x^3+6xy-4y^2)^2
no x=2 is not a function, as there is nothing that's changing.
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
The
i mean can we replace the roles of x and y and call x=2 as function
The
if x =y and x = 800 what is y
y=800
800
Bg
how do u factor the numerator?
Nonsense, you factor numbers
Antonio
You can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2
The
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
Antonio
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
volume between cone z=√(x^2+y^2) and plane z=2
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio
How do we find the horizontal asymptote of a function using limits?
Easy lim f(x) x-->~ =c
Antonio
solutions for combining functions
what is a function? f(x)
one that is one to one, one that passes the vertical line test
Andrew
It's a law f() that to every point (x) on the Domain gives a single point in the codomain f(x)=y
Antonio
is x=2 a function?
The
restate the problem. and I will look. ty
is x=2 a function?
The
What is limit
it's the value a function will take while approaching a particular value
Dan
don ger it
Jeremy
what is a limit?
Dlamini
it is the value the function approaches as the input approaches that value.
Andrew
Thanx
Dlamini
Its' complex a limit It's a metrical and topological natural question... approaching means nothing in math
Antonio
is x=2 a function?
The
3y^2*y' + 2xy^3 + 3y^2y'x^2 = 0 sub in x = 2, and y = 1, isolate y'
what is implicit of y³+x²y³=5 at (2,1)
tel mi about a function. what is it?
Jeremy
A function it's a law, that for each value in the domaon associate a single one in the codomain
Antonio
function is a something which another thing depends upon to take place. Example A son depends on his father. meaning here is the father is function of the son. let the father be y and the son be x. the we say F(X)=Y.
Bg
yes the son on his father
pascal
a function is equivalent to a machine. this machine makes x to create y. thus, y is dependent upon x to be produced. note x is an independent variable
moe
x or y those not matter is just to represent.
Bg