# 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 $f\left(x\right)\text{:}$

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

What is derivative of antilog x dx ?
what's the meaning of removable discontinuity
what's continuous
Brian
an area under a curve is continuous because you are looking at an area that covers a range of numbers, it is over an interval, such as 0 to 4
Lauren
using product rule x^3,x^5
Kabiru
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Sunny
Luke 17:21 nor will they say, See here or See there For indeed, the kingdom of God is within you. You've never 'touched' anything. The e-energy field created by your body has pushed other electricfields. even our religions tell us we're the gods. We live in energies connecting us all. Doa/higgsfield
Scott
if you have any calculus questions many of us would be happy to help and you can always learn or even invent your own theories and proofs. math is the laws of logic and reality. its rules are permanent and absolute. you can absolutely learn calculus and through it better understand our existence.
Scott
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Bilal
help the integral of x^2/lnxdx
Levis
also find the value of "X" from the equation that follow (x-1/x)^4 +4(x^2-1/x^2) -6=0 please guy help
Levis
Use integration by parts. Let u=lnx and dv=x2dx Then du=1xdx and v=13x3. ∫x2lnxdx=13x3lnx−∫(13x3⋅1x)dx ∫x2lnxdx=13x3lnx−∫13x2dx ∫x2lnxdx=13x3lnx−19x3+C
Bilal
itz 1/3 and 1/9
Bilal
now you can find the value of X from the above equation easily
Bilal
Pls i need more explanation on this calculus
usman
usman from where do you need help?
Levis
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Levis
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yh
Gbesemete
How do i differentiate between substitution method, partial fraction and algebraic function in integration?
usman
you just have to recognize the problem. there can be multiple ways to solve 1 problem. that's the hardest part about integration
Lauren
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we asking the question cause only the question will tell us the right answer
Sunny
find integral of sin8xcos12xdx
don't share these childish questions
Bilal
well find the integral of x^x
Levis
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Levis
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Bilal
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Levis
x^x ... no anti derivative for this function... but we can find definte integral numerically.
Bilal
thank you Bilal Kumhar then how we may find definite integral let say x^x,3,5?
Levis
evaluate 5-×square divided by x+2 find x as limit approaches infinity
i have not understood
Leo
Michael
welcome
Sunny
I just dont get it at all...not understanding
Michagaye
0 baby
Sunny
The denominator is the aggressive one
Sunny
wouldn't be any prime number for x instead ?
Harold
or should I say any prime number greater then 11 ?
Harold
just wondering
Harold
I think as limit Approach infinity then X=0
Levis
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Leonito
Fond the value of the six trigonometric function of an angle theta, which terminal side passes through the points(2x½-y)²,4
What's f(x) ^x^x
What's F(x) =x^x^x
Emeka
are you asking for the derivative
Leo
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Levis
iam sorry f(x)=x^x it means the output(range ) depends to input(domain) value of x by the power of x that is to say if x=2 then x^x would be 2^2=4 f(x) is the product of X to the power of X its derivatives is found by using product rule y=x^x introduce ln each side we have lny=lnx^x =lny=xlnx
Levis
the derivatives of f(x)=x^x IS (1+lnx)*x^x
Levis
what is a maximax
A maxima in a curve refers to the maximum point said curve. The maxima is a point where the gradient of the curve is equal to 0 (dy/dx = 0) and its second derivative value is a negative (d²y/dx² = -ve).
Viewer
what is the limit of x^2+x when x approaches 0
it is 0 because 0 squared Is 0
Leo
0+0=0
Leo
simply put the value of 0 in places of x.....
Tonu
the limit is 2x + 1
Nicholas
the limit is 0
Muzamil
limit s x
Bilal
The limit is 3
Levis
Leo we don't just do like that buddy!!! use first principle y+∆y=x+∆x ∆y=x+∆x-y ∆y=(x+∆x)^2+(x+∆x)-x^2+x on solving it become ∆y=3∆x+∆x^2 as ∆x_>0 limit=3 if you do by calculator say plugging any value of x=0.000005 which approach 0 you get 3
Levis
find derivatives 3√x²+√3x²
3 + 3=6
mujahid
How to do basic integrals
the formula is simple x^n+1/n+1 where n IS NOT EQUAL TO 1 And n stands for power eg integral of x^2 x^2+1/2+1 =X^3/3
Levis
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find the integral of tan tanxdx
-ln|cosx| + C
Jug
lnSecx+c
Levis