Simplifying Derivatives
In this tutorial kit the goal is to show you how to
simplify a derivative until it is in its simplest form. But
before we do that we are going to have to teach you the
basics of derivatives.
Lets start by learning what a derivative is.
The derivative of a function is another function. The one
whose value at a point is equal to the rate of change of
the given. By rate of change we mean the instantaneous rate
of change at a particular point rather than the average
rate. In terms of the graph, the instantaneous rate of
change corresponds to the slope of the tangent line.
Instantaneous rates are basically just the limit of the
average rate.
The diagram below shows a picture of what you would see for
the slope of a tangent line.
For any function, f, its derivative is the function, f ',
whose value at any point x is given by
F ' (x) = lim
H->0
f(x+h) - f(x)
H
In the above Equation the symbol h refers to a quantity
that is being taken to a particular limiting value (in this
case 0), and since h is not a value on which the limit
depends it could be replaced by any other symbol without
affecting the result. Also the values of h can be either
positive or negative.
So I guess by now you are probably wondering well what does
the derivative tell us about a function.



A positive derivative means that the function is
increasing
A negative derivative means that the function is
decreasing
A zero derivative means that the function has some
special behavior at the given point, like a max or a
min.
Now that you have learned what a derivative is we are going
to go through the 5 rules with you.
Rule # 1: the sum rule or power rule
Derivative = m tan
Y’
Dy/dx
derivatives
F’ (x)
* all of these represent
When doing derivatives there are a few things you will need
to know:





If y= c (constant) the y’ = 0 (y’ is y prime or the 1st
derivative), DxY= 0 (The derivative of Y with respect
to X), DY/DX = 0, and F’ (X) = 0 (no growth)
If y= mx + b the y’ = m, DY/DX = M (y grows with
respect to x at the rate of m), f’= m, DxY = m.
When doing derivatives there can be no negative
exponents, in order to get rid of these you are going
to have to rationalize, don’t worry there will be a
section in the kit that will tell you how.
There can be no fractional exponents
And finally there can be no radicals in the
denominator.
Ok now finally we’re going to show you how to do
derivatives.
Here’s an example and beside it will be steps explaining
how to do it.
Y= X2+ x1 + 3
one from exponent
Take exponent down & subtract
Y’= 2x + 1
is 0
Remember the derivative of a constant
The y1 represents velocity, instantaneous growth rate, and
slope of a tangent and the slope of a curve. It is shown by
DY/DX or DxY or F’ (x)
Since this can be simplified even more you have to know
that after y’ is y’’ and the y’’’ and so on.
So y’’= 2
The y’’ represents acceleration it is shown by D2Y/Dx2 or
Dx2Y or F’’ (X).
The 2’s in the D2Y0/Dx2 are simply telling you it is the 2nd
derivative.
Now on to the next rule
Rule # 2: product rule
Ok so here’s how you do it:
Take the derivative of the 1st bracket and multiply by the
second bracket. Add to this derivative of the second
bracket multiplied by the first bracket.
d
dx
[f (x) g (x)] = f (x) d [g (x)] + g (x) d [f (x)]
dx
dx
(Derivative 1st X 2nd) + (derivative 2nd X 1st)
Here’s an example that will show you how to do it:
Y= (2x-1)(4x2-6x-3)
Y= [(2)(4x2-6x-3)] +[(8x-6)(2x-1)] in the 1st bracket is
derivative of 1st X 2nd and in the 2nd bracket is the
derivative of the 2nd X 1st.
Now all you have to do is combine the brackets and then like
terms.
Y=8x2-12x-6+16x2-20x+6
Y=24x2-32x
Later on in the portfolio we will have examples showing you
how to do it then there will be some examples for you to try
your self.
Ok hopefully you understand so Far..
Rule # 3: Quotient Rule
*Ok since it is the quotient rule that means fractions
So take the derivative of the numerator and multiply it by
the denominator. Then subtract from this the derivative of
the denominator multiplied by the numerator. This will all
be over the denominator squared.
In case you have forgotten the numerator is the top numbers
and the denominator is the bottom.
D [n(x)] = d(x) d/dx [n(x)] – n(x) d/dx [d(x)]
Dx d(x)
[d (x)]2
Or in the form of words:
(Derivative of numerator X denominator) – (derivative of the denominator
X numerator)
(Denominator)2
Ok so here’s an example that should help you better
understand what’s going on.
Y= x2+x+1
X2-x-1
Y=[(2x+1)(x2-x-1)]-[(2x-1)(x2+x+1)]
(X2-x-1)2
3
2
Y=[2X -2X -2X-X2-X-1]-[2X3+X2+X-1]
*What we did here was
combine the brackets
(x2-x-1)2
Now we will combine
like terms
Y=(2x3-x2-3x-1)-(2x3+x2+x-1)
* now we combine the
brackets again
(x2-x-1)2
Y=
-2x2-4x
* and finally we can simplify
and we are done
(x2-x-1)2
y=-2x(x+2)
(x2-x-1)2
Ok now we’re almost done with all the rules here comes rule
# 4
Rule # 4: The Chain Rule
Dy = dy X du
Dx
du
dx
(Derivative with respect to the bracket) X (derivative of what’s
inside the bracket)
Ok here’s an example with steps to show you how to do it:
Y= (-x2+4x-2)2
Y= [2(-x2+4x-2)][(-2x+4)] the 1st bracket is the derivative with
respect to the bracket and the 2nd bracket it the derivative of
what’s inside. Now all you have to do is simplify.
Y= (-4x+8)(-x2+4x-2)
All right we are finally at the last rule
Rule # 5: Implicit Differentiation
I guess this isn’t really a rule but it is important to
know, so lets call it one anyway.
What this system does is it finds dy/dx without first
solving explicitly in a relation not in the form y= f(X)
We are trying to isolate dy/dx
So that may sound a little confusing so here’s an example
showing you just how its done:
X2+y2=25 (don’t forget the derivative of a constant is 0)
2x+2ydy =0 here the y will become dy/dx (y’)
dx
2ydy=-2x
dx
dy=-2x/2y
dx
dy=-x/y
dx
So there are the five rules of derivatives
Just incase you don’t understand here are a few more examples of
the rules.
Examples:
Sum rule:
# 1 y= x-10
y=-10x-11
#2
y=3x3+2x2
y’=9x2+4x
y’’= 18x+4
Product Rule:
#1
y=[(1+x)(1+2x)][(1+3x)]
y=[(2)(1+x)+(1)(1+2x)][(1+3x)+3(1+x)(1+2x)]
y=[(2+2x+1+2x][(1+3x)+(3+3x)(1+2x)
y=(3+4x)(1+3x)+(3+6x+3x+6x2)
y=(3+9x+4x+12x2)+(3+6x+3x+6x2)
y=(12x2+13x+3)+(6x2+9x+3)
y=(18x2+22x+6)
#2
y+(3x-4)(x+7)
y=39-x+7)+-1(3x+4)
y=-3x+21-3x+4
y=-6x+25
Quotient Rule:
#1
y=1-x2
1+x2
y=(-2x)(1+x)-(2x)(1-x2)
(1+x)
*Continue of example # 1
y=(-2x-2x3)-(2x-2x3)
(1+x)
y=-2x3-2x-2x+2x3
(1+x)
y= -4x
(1+x)
#2
y=x5-x3+1
4x
y=x4 – x2 + x-1
4
4
4
y=x3- x- 1
2 4x2
5
y=4x -2x3-1
4x2
Chain rule:
#1
y=(1+x2)-10
y=-10(1+x)-11 * (3x2)
y=-30x2(1+x)-11
#2
y= (3x5-x3/2)4
y= 4(3x5-x3/2)(15x4-3/2x1/2)
y=(60x4-6x)(3x5-x3/2)3
Implicit Differentiation
#1
x-y=(x+y)2
x-y+x2+2xy+y2
1-dy/dx=2x+2y+2xdy/dx+2ydy/dx
1-2x-2y=(2x+2y+1) dy/dx
1-2x-2y-(1+2x+2y)=dy/dx
#2
3 = 2x2 + 4xy + 6y3.
dy/dx(3) = d/dx(2x2 + 4xy + 6y3)
0 = 4x + 4y + 4x(dy/dx) + 18y2(dy/dx)
(4x + 18y2) dy/dx = -4x-4y
dy/dx = (-4x-4y)/(4x + 18y2)
Ok so those weren’t to bad, now here’s a twist combine the rules
together in a question, it’s not much harder all you have to do
it take it one step at a time making sure to always follow the
rules for each one.
#1
y=(2x-1)3(2x+2)4 ok so here we are going to have to use the
chain rule for the brackets and the product rule for the
entire equation.
Y=[3(2x-1)2(2)](2x+2)4=[4(2x+2)3(2)](2x-1)3
Y=[6(2x-1)2](2x+2)4+[8(2x+2)3](2x-1)3
Y=(2x-1)2+(2x+2)3[6(2x+2)(1)+8(2x-1)(1)]
Y=(2x-1)2+(2x+2)3[12x+12+16x-8]
Y=(2x-1)2+(2x+2)3[28x+4]
Y=(2x-1)2+(2x+2)3[4(7x+1)]
#2
d(x2y-2y2x)5
ok this is a mix of implicit differentiation
and the quotient
dx
rule
5(x2y-2y2x)4 (2xy+x2 dy/dx-2y2-4xy dy/dx)
5(x2y-2y2x)4 [(2xy-2y2+x2-4xy)]
Ok so now that we understand how to do derivatives we need to
understand how to simplify them. Which means we have to follow 3
very important rules:
1) no fractional exponents
2) no negative exponents
3) no radicals in the denominator
We’ll start with # 1
No fractional exponents
So just how to we do this?? Well it involves changing the base as
well as the exponent
For example if we had x1/2 it would change to x this is because
to the power of ½ is the same as taking the square root of
something. Now that was pretty simple right?
ok so here are some more examples of how to get rid of the
exponents
*41/3 =34
*62/3 =(61/3)2 =(36)2
*27/3=22+1/3=22*21/3=432
Ok so those were just examples oh how to get rid of the
fractions now lets try it with some derivatives.
#1
y=6x2/3
y=4x-1/3
y= 4
x1/3
y= 4* 3x2
3x
3x2
y=4(3x2)
x
#2
y= 4x3/4 – x-3/4
y=3x-1/4 + 3/4x-7/4
y= (3/x)1/4 + 3/4x7/4
y= 3(4x) + ¾ 4x7
y= 3(4x3/x )+ ¾(x4x3)
y=3(4x3/x) + 3(4x/4x2)
y= 12(4x3 )+3(4x)
4x2
Ok so now that we have done that rule let’s go on to #2
No negative Exponents
x-n = 1 / xn
So we see by this equation that when we have a negative exponent
we take the base and exponent change the negative to a positive
and divide it by 1
Ok so here are some basic examples showing you how this works
Oh and just so you know decimals are the same as fractions we
need to change them to whole numbers.
*3-2=1/32=1/9
*10-2=1/100
ok now lets try a couple examples with derivatives
#1
y=x(x-3)2 y(xy-2)-3
(y2)3 y-3(x2)3
y= x(x-6) y (x-3)y6
y6 y-3 x6
y= x(x-6) y(x-3) y6 * y-6 y3 x-6
y= x (x-6) (x-3) (x-6) * y (y6) (y-6) (y3)
y= x-14 y4
y=y4
x14
y=( 2x-1 )2
2x
y=4x2-4x-1
4x2
y=1-x-1+1/4x-2
y=x2-1/2x-3
y=1/x2-1/2x3
y=2x-1
2x3
#2
ok and finally we have come to the last one, rule #3
No radicals in the denominator
It’s actually quite simple all you do is all we have to do is
multiply the radical by itself and then we multiply the numerator
by the same number. This is only in the case that you would have
only one in the denominator.
Lets say you have 2two radicals in the denominator well you
would Multiply top and bottom by the difference if the
original is a sum and the sum if the original is a
difference.
Here are a few examples so that you understand. Then we’ll show
you some with derivatives.



11/6= 11/6=116/66 = 6*6/6
1/52 = 1/52 * 516/516 = 516/2
a + b = (a + b) * (a + b) = (a + b)2 = a+2 (ab) + b
a - b
(a - b)
(a + b)
a – b
a – b
so now that we’ve seen how to get rid of radicals lets try some
with derivatives.
#1
y= 2x2/3 – 7x 7/2
y=4/7x-5/7 – 49/2(x5/2)
y= 4/7 – 49/2(2x5)
x5/7
y=4/7 * 7x2 – 49/2 (2x5)
7x5
7x2
y= 4/7 (7x2) – 49/2(2x5)
x
#2
y= (25 – x2)1/2
y= ½ (25 – x2)-1/2 (-2x)
y= -1/x (25 – x2)-1/2
y=
-x *(25 – x2)
(25-x2) (25-x2)
y=-x(25-x2)
25 – x2
All right so finally we know just how to do derivatives and how
to rationalize them as well. Great, so now all we have to learn
is trig derivatives.
Trig Derivatives
Lets start by introducing you to what the derivatives of sin, cos
and tan are then we’ll do a few examples that you can follow so
you will hopefully understand
List of trig derivatives
* d (sinx)= cosx
dx
* d(cosx)= -sinx
dx
* d (tanx) = d(sinx) = sec2x
dx
dx(cosx)
* d(secx) = secxtanx
dx
* d(cotx) = -csc2x
dx
* d(cscx)= (-cscx)(cotx)
dx
Let’s go over some examples just so you understand what’s going
on. All of the rules still apply when doing trig but instead of
numbers you are using trig. It’s basically the same thing!
#1
#2
y= sin2x
y= sinx + xcosx
cotA + tanA= 2csc2A
cosA + sinA =
2_
sinA
cosA
sin2A
cos2A + sin2A =
2__
sinAcosA
2sinAcosA
1__ =
1___
sinAcosA
sinAcosA
Ok so hopefully our tutorial kit helped you. Now that we’re done
explaining everything we’re going to put in 15 example questions
for you to try and a little test at the back. Don’t worry there
will be an appendix at the back with the answers so you can check
and see how you’ve done. Also we’re going to put in some top websites and books that we have found that can help you out even
more.
Practice Problems
Solve and simplify
1) y = x24
11) y =
x_____
(2x-7)(x-7)
2) f(x)= 4x7
12) y = (x5 – 2)
x2 + 6x
3) y = t6 – 2t3 + 99
13) f(x)= ax + b
ax - b
4) y = 3x3 + x4 – 7
14) y = 2x2
5) y = x4/7 + 3x2/5
15) y = 5 – x2
2x + 3
6) y = x1/3 – x3/8
16) y = sin2x
7) y = x – 3x3
5
17) y = xsinx
8) y = x5 + 3x2
(6x2)2
18) y = cos2x
sin2x
9) y = (x3)2(x2)3
19) y = 2cosx
sin4x
10)y =
x-x2+1
20) y =_sin2A_
-sin4A
x
Test!
___
1)y=
(10x+9)2
(10x-9)2
2)_3x_ - 51+x
1+x
___
3)y= 1+1+x
4)y= [x+sin2x]3
5)y= 1+cos2x
1-cos2x
6)y= (5x6-4x)100
7)y= 4cosx4_
1-cosx3
8)y= 5x2-x3
9)y= 2x6
7x3
10)y= 5sinx3-cosx6
3tanx2
Apendix
Answers To the Practice Problems
1) y’= 24x23
2) f(x)’= 28x6
3) y’= 6x2(x3-1)
4) y’= x2(4x+9)
5) y’= 4x3 + 6x
6) y’= 1 – 3x2
3
8
7) y’= 1 –9x2
8) y’= 5x4 – 1
6x3
9) y’= 12x11
10)y’= -2x +1 –1
x2
11) y’= 2x2-14x+49
(7-2x)2
12) y’= 3x2+6 +4
x3
13)y’=
1-
b
a(x)2
a’(x)
14)y’= 22x–1+2
15)y’= -3x2
2
16)y’= 2sinxcosx
17)y’= sinx + xcosx
Continue of Answers
18)y’= -2cotx csc2x
19)y’= -(3cos(2x)+5)csc5x
20)y’= 2cotA csc2A
Answers to the Test
1)y’= -360(10x+9)
(10x-9)3
___
2)y’= 1+x(-2x-1)
2(1+x)2
3)y’= ¼ ___1______
1+1+x1+x
4)y’= 3(x+sin2x)2(sinx+cosx)2
5)y’= -4cosx
sin3x
6)y’= (3000x5-400)(5x6-4x)99
7)y’= (3-16cos3(x))sin(x)
8)y’= (10-3x)x
9)y’= 12x2
7
10)y’= -1/24(4cos(2x)+3cos(4x)-180sin(x)+1)sin
Bibliography
1) http://www.calc101.com/webMathematica/MSP/Calc101/WalkD
this site is great you can just go on and type in any derivative you
want them to do and they go through everything step by step.
2) http://www.langara.bc.ca/mathstats/resource/onWeb/calculus/derivatives/definition.htm
3)
http://ficp.engr.utexas.edu/calculus/module2.htm
4) Gage Calculus “Gary Flewelling, Wendy Warburton, Jack Weiner”
1989, Gage Educational Publishing Company.
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