# 3.3 Differentiation rules

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• State the constant, constant multiple, and power rules.
• Apply the sum and difference rules to combine derivatives.
• Use the product rule for finding the derivative of a product of functions.
• Use the quotient rule for finding the derivative of a quotient of functions.
• Extend the power rule to functions with negative exponents.
• Combine the differentiation rules to find the derivative of a polynomial or rational function.

Finding derivatives of functions by using the definition of the derivative can be a lengthy and, for certain functions, a rather challenging process. For example, previously we found that $\frac{d}{dx}\left(\sqrt{x}\right)=\frac{1}{2\sqrt{x}}$ by using a process that involved multiplying an expression by a conjugate prior to evaluating a limit. The process that we could use to evaluate $\frac{d}{dx}\left(\sqrt{x}\right)$ using the definition, while similar, is more complicated. In this section, we develop rules for finding derivatives that allow us to bypass this process. We begin with the basics.

## The basic rules

The functions $f\left(x\right)=c$ and $g\left(x\right)={x}^{n}$ where $n$ is a positive integer are the building blocks from which all polynomials and rational functions are constructed. To find derivatives of polynomials and rational functions efficiently without resorting to the limit definition of the derivative, we must first develop formulas for differentiating these basic functions.

## The constant rule

We first apply the limit definition of the derivative to find the derivative of the constant function, $f\left(x\right)=c.$ For this function, both $f\left(x\right)=c$ and $f\left(x+h\right)=c,$ so we obtain the following result:

$\begin{array}{cc}\hfill {f}^{\prime }\left(x\right)& =\underset{h\to 0}{\text{lim}}\frac{f\left(x+h\right)-f\left(x\right)}{h}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{c-c}{h}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{0}{h}\hfill \\ & =\underset{h\to 0}{\text{lim}}0=0.\hfill \end{array}$

The rule for differentiating constant functions is called the constant rule    . It states that the derivative of a constant function is zero; that is, since a constant function is a horizontal line, the slope, or the rate of change, of a constant function is $0.$ We restate this rule in the following theorem.

## The constant rule

Let $c$ be a constant.

If $f\left(x\right)=c,$ then ${f}^{\prime }\left(c\right)=0.$

Alternatively, we may express this rule as

$\frac{d}{dx}\left(c\right)=0.$

## Applying the constant rule

Find the derivative of $f\left(x\right)=8.$

This is just a one-step application of the rule:

${f}^{\prime }\left(8\right)=0.$

Find the derivative of $g\left(x\right)=-3.$

0

## The power rule

We have shown that

$\frac{d}{dx}\left({x}^{2}\right)=2x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\frac{d}{dx}\left({x}^{1\text{/}2}\right)=\frac{1}{2}{x}^{\text{−}1\text{/}2}.$

At this point, you might see a pattern beginning to develop for derivatives of the form $\frac{d}{dx}\left({x}^{n}\right).$ We continue our examination of derivative formulas by differentiating power functions of the form $f\left(x\right)={x}^{n}$ where $n$ is a positive integer. We develop formulas for derivatives of this type of function in stages, beginning with positive integer powers. Before stating and proving the general rule for derivatives of functions of this form, we take a look at a specific case, $\frac{d}{dx}\left({x}^{3}\right).$ As we go through this derivation, pay special attention to the portion of the expression in boldface, as the technique used in this case is essentially the same as the technique used to prove the general case.

## Differentiating ${x}^{3}$

Find $\frac{d}{dx}\left({x}^{3}\right).$

$\begin{array}{ccccc}\hfill \frac{d}{dx}\left({x}^{3}\right)& =\underset{h\to 0}{\text{lim}}\frac{{\left(x+h\right)}^{3}-{x}^{3}}{h}\hfill & & & \\ & =\underset{h\to 0}{\text{lim}}\frac{{x}^{3}+3{x}^{2}h+3x{h}^{2}+{h}^{3}-{x}^{3}}{h}\hfill & & & \begin{array}{c}\text{Notice that the first term in the expansion of}\hfill \\ {\left(x+h\right)}^{3}\phantom{\rule{0.2em}{0ex}}\text{is}\phantom{\rule{0.2em}{0ex}}{x}^{3}\phantom{\rule{0.2em}{0ex}}\text{and the second term is}\phantom{\rule{0.2em}{0ex}}3{x}^{2}h.\phantom{\rule{0.2em}{0ex}}\text{All}\hfill \\ \text{other terms contain powers of}\phantom{\rule{0.2em}{0ex}}h\phantom{\rule{0.2em}{0ex}}\text{that are two or}\hfill \\ \text{greater.}\hfill \end{array}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{3{x}^{2}h+3x{h}^{2}+{h}^{3}}{h}\hfill & & & \begin{array}{c}\text{In this step the}\phantom{\rule{0.2em}{0ex}}{x}^{3}\phantom{\rule{0.2em}{0ex}}\text{terms have been cancelled,}\hfill \\ \text{leaving only terms containing}\phantom{\rule{0.2em}{0ex}}h.\hfill \end{array}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{h\left(3{x}^{2}+3xh+{h}^{2}\right)}{h}\hfill & & & \text{Factor out the common factor of}\phantom{\rule{0.2em}{0ex}}h.\hfill \\ & =\underset{h\to 0}{\text{lim}}\left(3{x}^{2}+3xh+{h}^{2}\right)\hfill & & & \begin{array}{c}\text{After cancelling the common factor of}\phantom{\rule{0.2em}{0ex}}h,\text{the}\hfill \\ \text{only term not containing}\phantom{\rule{0.2em}{0ex}}h\phantom{\rule{0.2em}{0ex}}\text{is}\phantom{\rule{0.2em}{0ex}}3{x}^{2}.\hfill \end{array}\hfill \\ & =3{x}^{2}\hfill & & & \text{Let}\phantom{\rule{0.2em}{0ex}}h\phantom{\rule{0.2em}{0ex}}\text{go to 0.}\hfill \end{array}$

#### Questions & Answers

Good morning,,, how are you
Harrieta Reply
d/dx{1/y - lny + X^3.Y^5}
mogomotsi Reply
How to identify domain and range
Umar Reply
hello
Akpevwe
He,,
Harrieta
hi
Dr
hello
velocity
I only talk to girls
Dr
women are smart then guys
Dr
Smarter
Adri
sorry
Dr
hi adri ana
Dr
:(
Shun
was up
Dr
hello
Adarsh
is it chatting app?.. I do not see any calculus here. lol
Adarsh
Find the arc length of the graph of f(x) = In (sinx) on the interval [Π/4, Π/2].
mukul Reply
Sand falling freely from a lorry form a conical shape whose height is always equal to one-third the radius of the base. a. How fast is the volume increasing when the radius of the base is (1m) and increasing at the rate of 1/4cm/sec Pls help me solve
ade
show that lim f(x) + lim g(x)=m+l
BARNABAS Reply
list the basic elementary differentials
Chio Reply
Differentiation and integration
Okikiola Reply
yes
Damien
proper definition of derivative
Syed Reply
the maximum rate of change of one variable with respect to another variable
Amdad
terms of an AP is 1/v and the vth term is 1/u show that the sum of uv terms is 1/2(uv+1)
Inembo Reply
what is calculus?
BISWAJIT Reply
calculus is math that studies the change in math, such as the rate and distance,
Tamarcus
what are the topics in calculus
Augustine
what is limit of a function?
Geoffrey Reply
what is x and how x=9.1 take?
Pravin Reply
what is f(x)
Inembo Reply
the function at x
Marc
also known as the y value so I could say y=2x or f(x)= 2x same thing just using functional notation your next question is what is dependent and independent variables. I am Dyslexic but know math and which is which confuses me. but one can vary the x value while y depends on which x you use. also
Marc
up domain and range
Marc
enjoy your work and good luck
Marc
I actually wanted to ask another questions on sets if u dont mind please?
Inembo
I have so many questions on set and I really love dis app I never believed u would reply
Inembo
Hmm go ahead and ask you got me curious too much conversation here
Adri
am sorry for disturbing I really want to know math that's why *I want to know the meaning of those symbols in sets* e.g n,U,A', etc pls I want to know it and how to solve its problems
Inembo
and how can i solve a question like dis *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
next questions what do dy mean by (A' n B^c)^c'
Inembo
The sets help you to define the function. The function is like a magic box where you put inside stuff(numbers or sets) and you get out the stuff but in different shapes (forms).
Adri
I dont understand what you wanna say by (A' n B^c)^c'
Adri
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Adri
Ok so the set is formed by vectors and not numbers
Adri
A vector of length n
Adri
But you can make a set out of matrixes as well
Adri
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
Wait what's your math level?
Adri
High-school?
Adri
yes
Inembo
am having big problem understanding sets more than other math topics
Inembo
So f:R->R means that the function takes real numbers and provides real numer. For ex. If f(x) =2x this means if you give to your function a real number like 2,it gives you also a real number 2times2=4
Adri
pls answer this question *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
If you have f:R^n->R^n you give to your function a vector of length n like (a1,a2,...an) where all a1,.. an are reals and gives you also a vector of length n... I don't know if i answering your question. Otherwise on YouTube you havr many videos where they explain it in a simple way
Adri
I would say 24
Adri
Offer both
Adri
Sorry 20
Adri
Actually you have 40 - 4 =36 who offer maths or physics or both.
Adri
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
Adri
56-36=20 who give both courses... I would say that
Adri
solution: In a question involving sets and Venn diagram, the sum of the members of set A + set B - the joint members of both set A and B + the members that are not in sets A or B = the total members of the set. In symbolic form n(A U B) = n(A) + n (B) - n (A and B) + n (A U B)'.
Mckenzie
In the case of sets A and B use the letters m and p to represent the sets and we have: n (M U P) = 40; n (M) = 24; n (P) = 32; n (M and P) = unknown; n (M U P)' = 4
Mckenzie
Now substitute the numerical values for the symbolic representation 40 = 24 + 32 - n(M and P) + 4 Now solve for the unknown using algebra: 40 = 24 + 32+ 4 - n(M and P) 40 = 60 - n(M and P) Add n(M and P), as well, subtract 40 from both sides of the equation to find the answer.
Mckenzie
40 - 40 + n(M and P) = 60 - 40 - n(M and P) + n(M and P) Solution: n(M and P) = 20
Mckenzie
thanks
Inembo
Simpler form: Add the sums of set M, set P and the complement of the union of sets M and P then subtract the number of students from the total.
Mckenzie
n(M and P) = (32 + 24 + 4) - 40 = 60 - 40 = 20
Mckenzie
how do i evaluate integral of x^1/2 In x
ayo Reply
first you simplify the given expression, which gives (x^2/2). Then you now integrate the above simplified expression which finally gives( lnx^2).
Ahmad
by using integration product formula
Roha
find derivative f(x)=1/x
Mul Reply
-1/x^2, use the chain rule
Andrew
f(x)=x^3-2x
Mul
what is domin in this question
noman
all real numbers . except zero
Roha
please try to guide me how?
Meher
what do u want to ask
Roha
?
Roha
the domain of the function is all real number excluding zero, because the rational function 1/x is a representation of a fractional equation (precisely inverse function). As in elementary mathematics the concept of dividing by zero is nonexistence, so zero will not make the fractional statement
Mckenzie
a function's answer/range should not be in the form of 1/0 and there should be no imaginary no. say square root of any negative no. (-1)^1/2
Roha
domain means everywhere along the x axis. since this function is not discontinuous anywhere along the x axis, then the domain is said to be all values of x.
Andrew
Derivative of a function
Waqar
right andrew ... this function is only discontinuous at 0
Roha
of sorry, I didn't realize he was taking about the function 1/x ...I thought he was referring to the function x^3-2x.
Andrew
yep...it's 1/x...!!!
Roha
true and cannot be apart of the domain that makes up the relation of the graph y = 1/x. The value of the denominator of the rational function can never be zero, because the result of the output value (range value of the graph when x =0) is undefined.
Mckenzie
👍
Roha
Therefore, when x = 0 the image of the rational function does not exist at this domain value, but exist at all other x values (domain) that makes the equation functional, and the graph drawable.
Mckenzie
👍
Roha
Roha are u A Student
Lutf
yes
Roha

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