# 3.3 Differentiation rules  (Page 4/14)

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

By adding and subtracting $f\left(x\right)g\left(x+h\right)$ in the numerator, we have

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

After breaking apart this quotient and applying the sum law for limits, the derivative becomes

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

Rearranging, we obtain

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

By using the continuity of $g\left(x\right),$ the definition of the derivatives of $f\left(x\right)$ and $g\left(x\right),$ and applying the limit laws, we arrive at the product rule,

${j}^{\prime }\left(x\right)={f}^{\prime }\left(x\right)g\left(x\right)+{g}^{\prime }\left(x\right)f\left(x\right).$

## Applying the product rule to constant functions

For $j\left(x\right)=f\left(x\right)g\left(x\right),$ use the product rule to find ${j}^{\prime }\left(2\right)$ if $f\left(2\right)=3,{f}^{\prime }\left(2\right)=-4,g\left(2\right)=1,$ and ${g}^{\prime }\left(2\right)=6.$

Since $j\left(x\right)=f\left(x\right)g\left(x\right),{j}^{\prime }\left(x\right)={f}^{\prime }\left(x\right)g\left(x\right)+{g}^{\prime }\left(x\right)f\left(x\right),$ and hence

${j}^{\prime }\left(2\right)={f}^{\prime }\left(2\right)g\left(2\right)+{g}^{\prime }\left(2\right)f\left(2\right)=\left(-4\right)\left(1\right)+\left(6\right)\left(3\right)=14.$

## Applying the product rule to binomials

For $j\left(x\right)=\left({x}^{2}+2\right)\left(3{x}^{3}-5x\right),$ find ${j}^{\prime }\left(x\right)$ by applying the product rule. Check the result by first finding the product and then differentiating.

If we set $f\left(x\right)={x}^{2}+2$ and $g\left(x\right)=3{x}^{3}-5x,$ then ${f}^{\prime }\left(x\right)=2x$ and ${g}^{\prime }\left(x\right)=9{x}^{2}-5.$ Thus,

${j}^{\prime }\left(x\right)={f}^{\prime }\left(x\right)g\left(x\right)+{g}^{\prime }\left(x\right)f\left(x\right)=\left(2x\right)\left(3{x}^{3}-5x\right)+\left(9{x}^{2}-5\right)\left({x}^{2}+2\right).$

Simplifying, we have

${j}^{\prime }\left(x\right)=15{x}^{4}+3{x}^{2}-10.$

To check, we see that $j\left(x\right)=3{x}^{5}+{x}^{3}-10x$ and, consequently, ${j}^{\prime }\left(x\right)=15{x}^{4}+3{x}^{2}-10.$

Use the product rule to obtain the derivative of $j\left(x\right)=2{x}^{5}\left(4{x}^{2}+x\right).$

${j}^{\prime }\left(x\right)=10{x}^{4}\left(4{x}^{2}+x\right)+\left(8x+1\right)\left(2{x}^{5}\right)=56{x}^{6}+12{x}^{5}.$

## The quotient rule

Having developed and practiced the product rule, we now consider differentiating quotients of functions. As we see in the following theorem, the derivative of the quotient is not the quotient of the derivatives; rather, it is the derivative of the function in the numerator times the function in the denominator minus the derivative of the function in the denominator times the function in the numerator, all divided by the square of the function in the denominator. In order to better grasp why we cannot simply take the quotient of the derivatives, keep in mind that

$\frac{d}{dx}\left({x}^{2}\right)=2x,\text{not}\phantom{\rule{0.2em}{0ex}}\frac{\frac{d}{dx}\left({x}^{3}\right)}{\frac{d}{dx}\left(x\right)}=\frac{3{x}^{2}}{1}=3{x}^{2}.$

## The quotient rule

Let $f\left(x\right)$ and $g\left(x\right)$ be differentiable functions. Then

$\frac{d}{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{\frac{d}{dx}\left(f\left(x\right)\right)·g\left(x\right)-\frac{d}{dx}\left(g\left(x\right)\right)·f\left(x\right)}{{\left(g\left(x\right)\right)}^{2}}.$

That is,

$\text{if}\phantom{\rule{0.2em}{0ex}}j\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)},\text{then}\phantom{\rule{0.2em}{0ex}}{j}^{\prime }\left(x\right)=\frac{{f}^{\prime }\left(x\right)g\left(x\right)-{g}^{\prime }\left(x\right)f\left(x\right)}{{\left(g\left(x\right)\right)}^{2}}.$

The proof of the quotient rule    is very similar to the proof of the product rule, so it is omitted here. Instead, we apply this new rule for finding derivatives in the next example.

## Applying the quotient rule

Use the quotient rule to find the derivative of $k\left(x\right)=\frac{5{x}^{2}}{4x+3}.$

Let $f\left(x\right)=5{x}^{2}$ and $g\left(x\right)=4x+3.$ Thus, ${f}^{\prime }\left(x\right)=10x$ and ${g}^{\prime }\left(x\right)=4.$ Substituting into the quotient rule, we have

${k}^{\prime }\left(x\right)=\frac{{f}^{\prime }\left(x\right)g\left(x\right)-{g}^{\prime }\left(x\right)f\left(x\right)}{{\left(g\left(x\right)\right)}^{2}}=\frac{10x\left(4x+3\right)-4\left(5{x}^{2}\right)}{{\left(4x+3\right)}^{2}}.$

Simplifying, we obtain

${k}^{\prime }\left(x\right)=\frac{20{x}^{2}+30x}{{\left(4x+3\right)}^{2}}.$

Find the derivative of $h\left(x\right)=\frac{3x+1}{4x-3}.$

${k}^{\prime }\left(x\right)=-\frac{13}{{\left(4x-3\right)}^{2}}.$

It is now possible to use the quotient rule to extend the power rule to find derivatives of functions of the form ${x}^{k}$ where $k$ is a negative integer.

## Extended power rule

If $k$ is a negative integer, then

$\frac{d}{dx}\left({x}^{k}\right)=k{x}^{k-1}.$

## Proof

If $k$ is a negative integer, we may set $n=\text{−}k,$ so that n is a positive integer with $k=\text{−}n.$ Since for each positive integer $n,{x}^{\text{−}n}=\frac{1}{{x}^{n}},$ we may now apply the quotient rule by setting $f\left(x\right)=1$ and $g\left(x\right)={x}^{n}.$ In this case, ${f}^{\prime }\left(x\right)=0$ and ${g}^{\prime }\left(x\right)=n{x}^{n-1}.$ Thus,

questions solve y=sin x
Solve it for what?
Tim
you have to apply the function arcsin in both sides and you get arcsin y = acrsin (sin x) the the function arcsin and function sin cancel each other so the ecuation becomes arcsin y = x you can also write x= arcsin y
Ioana
what is the question ? what is the answer?
Suman
there is an equation that should be solve for x
Ioana
ok solve it
Suman
are you saying y is of sin(x) y=sin(x)/sin of both sides to solve for x... therefore y/sin =x
Tyron
or solve for sin(x) via the unit circle
Tyron
what is unit circle
Suman
a circle whose radius is 1.
Darnell
the unit circle is covered in pre cal...and or trigonometry. it is the multipcation table of upper level mathematics.
Tyron
what is function?
A set of points in which every x value (domain) corresponds to exactly one y value (range)
Tim
what is lim (x,y)~(0,0) (x/y)
limited of x,y at 0,0 is nt defined
Alswell
But using L'Hopitals rule is x=1 is defined
Alswell
Could U explain better boss?
emmanuel
value of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)
NIKI
can we apply l hospitals rule for function of two variables
NIKI
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
2x^3+6xy-4y^2)^2 solve this
femi
moe
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