Table of Contents
Introduction to Implicit Differentiation.....................…………………………………………………………Page 2
Analytical Example ………………………………………………………………………………………………………………Pages 2-3
Conceptual Example…………………………………………………………………………………………………………….Pages 3
AP Level Multiple Choice……………………………………………………………………………………………………..Pages 4-6
Calculator Problem……………………………………………………………………………………………………………..Pages 6-7
AP Level Free Response………………………………………………………………………………………………………Pages 7-10
Real Life Applications…………………………………………………………………………………………………………..Page 11
Mathematician Who Contributed to Implicit Differentiation........................................................Page 11
Integration by Parts Analytical Example………………………………………………………………………………..Page 12
Integration by Partial Fractions Analytical Example………………………………………………………………Page 13
Multiple Choice Partial Fractions………………………………………………………………………………………Page 14
Conceptual Problem Integration by Parts…………………………………………………………………………Page 15
Graphing Calculator for Partial Fractions………………………………………………………………………Page 16
AP Free Response for Integration by Parts…………………………………………………………………….Page 17
Real Life Applications for Integration by Parts and Partial Fractions…………………………………Page 18
Mathematician who Contributed to Integration by Parts and Partial Fractions……………….Page 18
Works Cited…………………………………………………………………………………………………………………….Page 19
By: Shaun Vadaketh and Matt Sorensen
Implicit Differentiation
Basically, implicit differentiation is used when you have an equation that has more than one ycomponent. For example, if you have a function like 2 y + y 2 = x 2 , you can’t really use normal
differentiation unless you simplify this function to a ‘y=…’ form. And that’s pretty inconvenient if you ask
me, so let’s NOT go down that route. So instead of going through the tedious process of isolating the ‘y’
and then using normal differentiation, we can use implicit differentiation.
Analytical Example: Differentiate 2 y  y 2  x 2
The first thing you should notice is that you can’t just do this problem by using normal differentiation:
there’s more than one ‘y’ component, meaning you’re not just deriving the x components – you also
have to derive the ‘y’ components too because they’re more than just a single ‘y’ (i.e. 2y and y 2 as
opposed to just y). That means that we’ve got to use implicit differentiation. Sounds complicated, but
don’t worry, IT ISN’T!
The way to think about implicit differentiation is to remember the derivative rules. Think back to when
you had simple equations like: y = 2x + 3. The way we derived an equation like that was by taking the
derivative of each term. So the derivative of ‘y’ is dy/dx. The derivative of 2x = 2 * 1 = 2. The 1 comes
from the fact that you have to take the derivative of ‘x’, which equals 1. THAT’S IMPORTANT! Remember
that when you take the derivative of anything that has a variable, you end with taking the derivative of
the variable. With ‘x’, it doesn’t really matter because the derivative of ‘x’ is just 1, but that isn’t
necessarily the case with ‘y’. To finish it off, the derivative of 3 = 0, so overall, if you derive y = 2x + 3,
you get: dy/dx = 2.
Okay, back to the original problem. Just like how we differentiated ‘2x’ above, we have to think of
anything ‘y-related’ as the derivative of the ‘y-related’ term multiplied by the derivative of y itself. The
derivative of y = dy/dx, so basically in implicit differentiation, you derive the ‘y-components’ normally
and just add dy/dx at the end. Confused? Trust me, it’s REALLY easy once you get the hang of it.
2y  y2  x2
The first thing you do is derive 2y. To do that, think of it like 2x: to derive 2x, it’s just 2 * 1 (the 2 comes
out and you take the derivative of x, which is just 1). So with 2y, the 2 comes out, and then you take the
derivative of y. In other words, it’ll be 2 *
dy
. Okay, that’s the first part.
dx
Next, we have to move on to the y 2 . Again, think of it like normal differentiation. Because of the power
rule, the exponent comes down and is multiplied by the term, and then the exponent is reduced by 1. In
other words, the 2 comes down and the exponent becomes 1 (because 2-1=1). That’ll get you 2y. Now
to finish it off, just like before, you have to take the derivative of y, which equals
dy
.
dx
dy
. The last part of the derivative is just deriving x 2 , and you do that the normal
dx
way we’ve always done it: we use the power rule. Thus, the derivative of x 2 is just 2x.
In all, you have 2y *
Overall, you have three separate derivatives. The derivative of 2y is 2 *
dy
. The derivative of y 2 is 2y *
dx
dy
. And the derivative of x 2 is 2x. Add those three up and the overall derivative of 2 y  y 2  x 2 is…
dx
(2 *
dy
dy
) + (2y *
) = 2x
dx
dx
Bang!
That’s just the basic outline. Now let’s move onto a conceptual problem so that you can get a better
understanding of how this stuff works. The entire premise of implicit differentiation comes from the
chain-rule, so let’s do a quick chain-rule problem to solidify your understanding of how implicit
differentiation came to be.
Given the function y  ( x 2  3) 4 , find the derivative of y with respect to x. What you would do is use
the normal chain rule like you always do: the exponent comes down and is multiplied by the term, then
the exponent is lessened by 1, and then the entire term is also multiplied by the derivative of the inside
term. Thus, you’d get: 4( x 2  3) 3 (2 x) . Now at this point you might be wondering what this has to do
with implicit differentiation.
Well, consider the x 2  3 as one big entity in and of itself. Treat it exactly like how you would if you had
something like y 2 and you were asked to differentiate that. See the similarity in how we differentiate?
With both terms, we use the full-out chain rule because they’re not just simple x’s—we can’t just ignore
the fact that their derivatives equal something OTHER than 1.
To put it simply, if you want to differentiate anything that has a y in it, use the same rules that you
would use if you were using the chain rule. Remember to take the derivative at the end!
Okay, now that we’ve gotten the hang of this, let’s try an AP-level multiple choice example. OH NO IT’S
SO HARD…not really.
Problem: If x 2  2 xy  3 y 2  8 , then
(A)
8  2 y  2x
6 y  2x
(B)
3y  x
yx
(C)
2x  2 y
6 y  2x
dy
=
dx
(D) 1/3
(E)
yx
3y  x
Right away, because there are multiple ‘x’ and ‘y’ terms and it’s asking for the derivative, you should
think “Oh, this is implicit differentiation!” And you’d be right, because that’s exactly what this is.
So to find
dy
, we obviously have to derive the equation. Just like before, we’ll take each term by itself
dx
and derive it individually, and then at the end we’ll add them up to get the entire derivative.
First up is x 2 . To derive x 2 , you use the power rule: the degree of the exponent is multiplied by the
term, and then the exponent is lessened by 1. Basically, the 2 comes down and is multiplied by the term,
and the degree of the exponent goes from 2 to 1. So overall, you’ll have 2 * ( x 21 ), which is equal to 2x.
Next up in the equation is ‘-2xy’. To derive this, we have to use a mix of the product rule and implicit
differentiation. Just like how if you have 2 ‘x’ terms, the product rule requires you to take the derivative
of the first term and multiply it by the second term, and then add that product to the derivative of the
second term multiplied by the first term. As with normal differentiation, the ‘-2’ can come out because
it’s an integer. So…
Overall, it will be: (2) * (( x *
dy
)  ( y * 1)) . The -2 is multiplied by the entire term because it came out
dx
in the beginning. The (x * dy/dx) comes from the first term (x) being multiplied by the second term’s
derivative (derivative of y = dy/dx). The (y * 1) comes from the second term (y) being multiplied by the
derivative of x (derivative of x = 1).
The third term in the equation is 3y 2 . To derive this, we use the power rule with the added mix of
implicit differentiation. Just like with the normal power rule, the exponent comes down and is multiplied
by the term—that means the 2 comes down and is multiplied to the ‘3y’. Then the exponent is
decreased by 1, which gives you a remaining power of 1. And lastly, as we did before with implicit
differentiation, we end things by taking the derivative of y, which equals dy/dx, and multiplying that to
the entire term. In all, it’ll look like: 6 y *
dy
.
dx
The last term in the equation is 8, and the derivative of 8 is just 0 because the derivative of any integer is
always 0.
OVERALL, we have four separate derivatives. Going through them quickly, the derivative of x 2  2 x , the
derivative of (-2xy) = (2) * (( x *
dy
dy
)  ( y * 1)) , the derivative of 3y 2 = 6 y *
, and the derivative of
dx
dx
8 = 0.
We add those up, and we get: 2x + (2) * (( x *
dy
dy
)  ( y * 1)) + 6 y *
= 0. To simplify that, I would
dx
dx
personally try to combine the terms in the parentheses to it’s not so cluttered and so that we can try to
get it to look like something that resembles the answer choices. If you distribute the (-2) to the terms
inside the parentheses, you should overall get: 2 x  2 x
dy
dy
 2y  6y
 0 . Now, at this point you’re
dx
dx
probably wondering, “We’re supposed to get dy/dx! Not some cluttered mess that has dy/dx mixed into
it!” EXACTLY!!
So to do that, we’ve got to factor out the dy/dx. The best way to do that is by isolating all the terms that
include dy/dx in them—that way we’ll be able to factor out the dy/dx from those terms. So, the terms
that DON’T have dy/dx in them are ‘2x’ and ‘-2y’. Move those onto the other side, so you have:
dy
dy
 6y
 2 y  2 x . Then, you have to factor out the dy/dx from the LEFT SIDE because that’s
dx
dx
dy
(2 x  6 y )  2 y  2 x . Then, you divide each side by (-2x
the side that has the dy/dx. That gets you:
dx
dy
2 y  2x

+ 6y), so that you can isolate the dy/dx. Overall, that should get you:
. Then we can
dx  2 x  6 y
 2x
simplify the right side of the equation by dividing the numerator and denominator by 2. That gets you:
dy
yx
. A simple cross-look match-up tells us that the only answer that matches that derivative

dx  x  3 y
is…choice E! Choice (E) is:
yx
, which is the same exact thing as what we got. So the answer is E.
3y  x
BANG!!
Now let’s do a problem where a calculator might be required. Usually with implicit differentiation,
there’s not much stuff to do with calculators because there’s no differentiation function on a calculator
to begin with, but sometimes if you are trying to find the derivative at a certain point, they can be useful
to plug in values.
Problem: Find
dy
, if x 2  y 3  10  5 xy at the point (2.3, 7).
dx
To do this, we first have to find the general derivative of the equation. Again, we’ll find the derivative of
each term individually and then add them up at the end. Makes things a lot easier in my opinion!
So the derivative of x 2 is found using the power rule—it’s equal to 2x (if you’re confused, I explained
the power rule briefly in the previous examples. The derivative of y 3 is found using the product rule and
implicit differentiation. Just like normal ‘product rule behavior’, the exponent comes down and is
multiplied by the overall term. So the 3 comes down. Then the exponent (3) is lessened by a degree of 1,
so it becomes 2. By this point, you should have: 3y 2 . That’s it, right? WRONG! Remember, whenever
you derive something that has a variable, always end by taking the derivative of that variable. Since
we’re deriving a y-term, we end by taking the derivative of y, which equals dy/dx, and then multiplying
that by the entire term. So in all, you should have: 3 y 2
dy
.
dx
The third term in the equation is ‘10’, and the derivative of that is just 0 because it’s an integer. The last
term is ‘-5xy’—OH MAN, ANOTHER PRODUCT RULE (UGH!!).
Just like before, with the product rule, we do it normally but make sure that when we take the
derivative of any ‘y’ term, we MAKE SURE TO TAKE THE DERIVATIVE OF Y.
With ‘-5xy’, the ‘-5’ can come out to the front because it’s an integer. Then we take the derivative of the
first term (x), which is just 1, and multiply that by the second term (y). Add that product to the derivative
of the second term (y), which is dy/dx, multiplied by the first term (x). In all, you should have:
(5) * ((1 * y )  (
 5 y  5x
dy
.
dx
dy
* x) . You can simplify that by distributing the (-5) to each term, so you’ll have:
dx
Overall, we have four separate terms, which when combined, should get you something like this:
2x  3y 2
dy
dy
 5 y  5 x . As with our AP multiple choice example, we have to try to get this in a
dx
dx
form where dy/dx is by itself. That means we have to isolate the terms containing dy/dx, and they are
3y 2
dy
dy
and  5 x
. Moving them to one side and the ‘2x’ and ‘-5y’ to the other side, you should get:
dx
dx
3y 2
dy
dy
dy
 5x
 5 y  2 x . Then factor out the dy/dx:
(3 y 2  5 x)  5 y  2 x . Divide each side
dx
dx
dx
by 3 y 2  5 x so that you can get dy/dx by itself, and you have…
dy  5 y  2 x
. WHEW!!!

dx 3 y 2  5 x
We’re not done yet though, so don’t get too excited! Since we’re trying to find the derivative (or dy/dx)
at the specific point (2.3, 7), we have to plug those coordinates into out general ‘dy/dx’ form to get the
exact solution. So we plug in 2.3 for our x’s and 7 for our y’s.
That will get us:
(5 * 7)  (2 * 2.3)
(3 * 7 2 )  (5 * 2.3)
USE YOUR CALCULATOR ON THE AP EXAM!! DON’T WASTE TIME.
Using your calculator, you should get dy/dx at the point (2.3,7) = -0.250
BANG!!!
Let’s move onto some more challenging stuff: AN AP LEVEL FREE RESPONSE!!
Problem: Consider the curve given by x 2  4 y 2  7  3xy
(a) Show that
dy 3 y  2 x

dx 8 y  3x
(b) Show that there is a point P with x-coordinate 3 at which the line tangent to the curve at P is
horizontal. Find the y-coordinate of P.
(c) Find the value of
d2y
at point P found in part (b). Does the curve have a local maximum, a
dx 2
local minimum, or neither at point P? Justify your answer.
Let’s start with part (b). JUST KIDDING…obviously you start with part (a) always.
Part (a) tells us to show that the derivative of the original equation equals that mess of a fraction. That
means that we have to solve for the derivative (making sure to show our work) and manipulate what we
get so that it looks like the derivative given in part (a).
To get the derivative, we’ll…AGAIN…take the derivative of each of the individual terms and then add
them up to get the entire derivative.
The first term is x 2 . To derive this, we use the power rule, which gets us 2x.
Next up is 4 y 2 . We have to use the power rule and implicit differentiation because there’s a y-term in
there. Because of the power rule, the exponent (2) comes down and is multiplied by the 4y, and then
the exponent itself is lessened by 1. Then we end by taking the derivative of y, which equals dy/dx, and
multiplying that by the entire term. In all, that gets us: (2 * 4) y 21 *
Simplified, that gets us: 8 y
dy
.
dx
dy
.
dx
The next term we have to derive is 7, and the derivative of 7 is 0 because it’s an integer (derivatives of
integers always equal 0).
The fourth and last term is 3xy. To derive this, we have to use the product rule and implicit
differentiation. To start, we can take the 3 out and multiply it by whatever derivative we get because it’s
an integer. That’ll get us: 3 * (the derivative of xy).
To derive xy, we use the product rule: we have to take the derivative of the first term (derivative of ‘x’ =
1) and multiply that by the second term (y), and then add that product to the product of the derivative
of the second term (derivative of ‘y’ = dy/dx) and the first term (x). Don’t forget that the 3 is multiplied
by that entire thing!
Overall, we have: 3 * ((1 * y )  ( x *
3 y  3x
dy
.
dx
dy
)) . We can simplify that by distributing the 3, and that gets us:
dx
We have four separate derivatives, and when we combine them in the format that’s given to us in the
original equation, we get:
2x  8 y
dy
dy
 3 y  3x .
dx
dx
Then we have to factor out the dy/dx, and we do that by isolating the terms that have dy/dx. That
means that we have to get all of the dy/dx terms on one side. In other words…
8y
dy
dy
 3x
 3 y  2 x . Then we have to factor out the dy/dx, which gives us:
dx
dx
dy
(8 y  3 x)  3 y  2 x . Lastly, we have to divide each side by (8y-3x) so that we can get dy/dx by
dx
itself.
IN ALL, that makes:
dy 3 y  2 x
. YES!! That’s good because that’s exactly what part (a) asked us to

dx 8 y  3x
prove was the derivative.
Next up is part (b): Show that there is a point P with x-coordinate 3 at which the line tangent to the
curve at P is horizontal. Find the y-coordinate of P.
Basically, the problem is asking us to demonstrate the fact that at (x=3), the derivative of the function is
0, and then it wants us to find the y-coordinate of the point at which (x=3) and the derivative is 0.
Remember, tangent lines being horizontal means that the DERIVATIVE is 0 at that point.
So…because it gives us the point (x=3), we basically have to plug x=3 into the derivative function and
show that the derivative at x=3 is 0. Again, our derivative function is:
dy 3 y  2 x
. Because there are

dx 8 y  3x
x’s and y’s in the general derivative function, we have to get a y-value too. So to figure out what the yvalue is at x =3, we have to plug x =3 into the original function and solve for y.
The original function is x 2  4 y 2  7  3xy . Plug in x = 3 into that equation, and you get:
9  4 y 2  7  9 y . We’ll have to use some factorization, so move everything onto one side.
Thus, you get: 4 y 2  9 y  2  0
Then factor that into two separate terms: (4 y  1)( y  2)  0 . Solving for y, you get: y 
1
,2 .
4
Because there’s two possible y-values for x = 3, we have to test each of them in the derivative and solve
for the one that gives us a derivative of 0. Let’s start with ¼.
Plugging in (3, ¼) into the derivative, we get:
Plugging in (3,2) into the derivative, we get:
.75  6
 .75  0
29
66
 0 . So, not only have we proved that there indeed
16  9
exists at point where x = 3 and the derivative is 0, but also we found the y-coordinate of that point,
which is 2. BANG!!!
Onwards!
Part (c): Find the value of
d2y
at point P found in part (b). Does the curve have a local maximum, a
dx 2
local minimum, or neither at point P? Justify your answer.
To find the value of the second derivative at a point, we have to find the general form of the second
derivative. To do that, we have to take the derivative of the derivative. Our original derivative equation
was:
dy 3 y  2 x
. To derive this, we have to use the quotient rule (OH NOOO!) and implicit

dx 8 y  3x
differentiation (because there’s y-terms in there). Using the quotient rule, we have:
(8 y  3x)(3
dy
dy
 2)  (3 y  2 x)(8  3)
dx
dx
. Point P, as we found in part (b), is: (3,2). So plugging that
2
(8 y  3x)
into our second derivative (and remember that for any dy/dx, we plug in the value of dy/dx at the point
(3,2)), we get:
(16  9)(0  2)  (6  6)(0  3)
. Overall, that equals -0.286.
(16  9) 2
Because at point P (3,2), dy/dx = 0 and
BANG!!!!
d2y
is negative, that means there is a local maximum at point P.
dx 2
At this point, you’re probably wondering what the point of implicit differentiation is. WELL,
young pupil, implicit differentiation is relevant to things we see EVERYDAY. For example, anything that
has to do with related rates has to do with implicit differentiation, because the way you solve a related
rates problem is through implicit differentiation. Imagine you have two planes traveling towards each
other, one going north and one going east. To figure out how fast the distance between the two planes
is changing, you have to use related rates! Now obviously the air-traffickers don’t have to use implicit
differentiation every time they want to direct the course of a plane, but they DO have machines that are
programmed to implicitly differentiate—convenient, right? Basically, anything that has to do with
changing distances/rates relative to another distance/rate has to do with implicit differentiation. And,
seeing as how many technological advancements require such knowledge, implicit differentiation is
VERY important in real life.
A mathematician who largely contributed to implicit differentiation was Leibniz. Although
Leibniz didn’t directly ‘create’ implicit differentiation, he was the first well-known individual to have
used to chain-rule, which is the primary component of implicit differentiation. Nobody really ‘created’
implicit differentiation, though—it’s one of those things that are pretty intuitive: it makes use of calculus
theorems to solve an ordinary function that is written in a weird form (hence the implicit differentiation
as opposed to the explicit differentiation). However, by creating the chain-rule, Leibniz essentially found
the easiest method to implicitly differentiate things because the chain-rule provides the framework for
how we differentiate. It dictates that whenever we take the derivative, we take the derivative of the
variable itself with respect to another variable—in most cases, we take the derivative of y with respect
to x. Ultimately, Leibniz handed down an absolutely beautiful ‘rule’ that allows us to implicitly
differentiate easily, so I personally credit him with most of the work surrounding implicit differentiation.
Integration by Parts
Integration by Parts is a theorem used in calculus to relate the integral of a product of functions
to the integral of their derivative and anti-derivative. Integration by parts transforms an integral
to make its solution easier to find.
General rule of Integration by Parts:
 udv  uv   vdu
Analytical example:
Solve:
 x cos xdx
At a first glance you may think to yourself that solving this is impossible.
Try this:
Set u  x and dv  cos xdx
Now you need to find du and v .
Since, u  x then by taking the derivative of both sides you find that du  dx .
Also, dv  cos xdx . By integrating both sides you find that v  sin x .
Now plug in to the General Rule for Integration by Parts:
x sin x   sin xdx
Then simplify.
x sin x  cos x  C
Integration by Partial Fractions
Integration using Partial Fraction Decomposition is a technique used to separate a
function that you cannot integrate into parts that you can easily integrate separately.
Here’s a problem that you can use Partial Fractions to Solve
At first, this integral is not easily solveable
x
1
dx
1
2
Using the Partial Fraction Technique we can separate it into parts using the polynomials factors.
1
A
B


x 1 x 1 x 1
2
Then multiply out and solve for A and B by plugging in values of x .
1  A( x  1)  B( x  1)
When x  1 , A  
When x  1, B 
1
2
1
2
Plug in your values for A and B .
1
1
1
 2  2
2
x 1 x 1 x 1

Now that you have the problem in a different form that you can integrate, solve it.
x
1
1
1
1
dx  

dx
2 x 1 x 1
1
2
1
ln x  1  ln x  1   C
2
1 x 1
 ln
C
2 x 1

Multiple Choice: Partial Fractions
1.
5x
 ( x  5)( x  8) dx 
25
1  ln x  5 
C
A) 
13  ln x  8 40 


D)
1
25 ln x  5  40 ln x  8   C
13
5 2
x
2
B)
C
1 3 3 2
x  x  40 x
3
2
E)
25
ln ( x  5)( x  8)  C
13
C)
1
25 ln x  5  40 ln x  8   C
13
Answer:
5x
 ( x  5)( x  8) dx 
5x
A
B


( x  5)( x  8) x  5 x  8
5 x  A( x  8)  B( x  5)
5(5)  A(5  8)
25
13
5(8)  B(8  5)
A
B
40
13
5x
25
1
40
1
 ( x  5)( x  8) dx  13  x  5 dx  13  x  8 dx
25
40
ln x  5  ln x  8  C
13
13
1
 25 ln x  5  40 ln x  8   C
13

Conceptual: Integration by Parts
Find du and v
 4x
2
e x dx
Answer:
Pick u and dv
u  4x 2
v?
du  ?
dv  e x dx
Solve for du and v
du  8x
v  ex
Graphing Calculator: Partial Fractions
Solve using a calculator.
7
x
2
3
1
dx 
 3x  10
Verify your answer using Partial Fractions.
Answer:
7
x
2
3
1
dx  0.172
 3x  10
1
A
B


x  3 x  10 x  5 x  2
2
1  A( x  2)  B( x  5)
A
B
1
7
1
7
7
7
1
1
1
1
3 x 2  3x  10 dx  7 3 x  2  x  5 dx
1 x2
ln
7 x5
 0.172
7

3
1 5 1 1 1 10
ln
 ln  ln
7 12 7 8 7 3
AP Free Response: Integration by Parts
Let f be the function defined for x  0, with f (e)  2 and f ' , the first derivative of f ,
4.
given by f ' ( x)  x 2 ln x.
(a) Write an equation for the line tangent to the graph of f at the point ( e , 2).
(b) Is the graph of f concave up or concave down on the interval 1  x  3 ? Give a
reason for your answer.
(c) Use antidifferentiation to find f (x ).
Answer to part (c):
 f ' ( x)dx   x
2
ln xdx
u  ln x
du 
dx
x
x3
3
dv  x 2 dx
v
x 3 ln x
x2
f ( x) 
  dx  C
3
3
f (x ) 
x 3 ln x x 3

C
3
9
f (e) 
e 3 ln e e 3
 C  2
3
9
C  2
3e 3 e 3
2e 3

 2
9
9
9
f (x ) 
x 3 ln x x 3
2e 3

2
3
9
9
Integration by parts and integration by partial fractions are simply ways to simplify the
integration process, so naturally, their real-life applications stem from the real-life applications of
integration itself. For example, calculating the solid of revolution of an object about a line requires
integration, and thus, depending on the equation that’s being integrated, you might have to use
integration by parts or partial fractions. Further uses include the generation of a general velocity
equation from an acceleration equation when given a ‘velocity-coordinate’—depending on the
acceleration equation, you may or may not have to use integration by parts or partial fractions
(remember, to get from acceleration to velocity, you have to integrate!). And these uses are so
important in the real world because they are used frequently in physics or mechanics: when
constructionists are trying to find the volume of a cylindrical basin, they could use integration to get a
very good approximate. Likewise, in real life if you can plot the acceleration of a car on a graph, it’s very
easy to get the velocity by using integration; and again, any integration applications have a lot to do with
partial fractions and integration by parts because they’re methods used to integrate.
The origins of integration by parts and partial fractions aren’t too well known because they’re
pretty intuitive mathematical techniques—they can be used in basic algebra as well as calculus, so their
origins definitely don’t come from something calculus-related (I’d expect them to have been created far
earlier than that). However, I think Sir Isaac Newton has to be mentioned. While he didn’t directly
contribute to the creation of the two techniques, he did create the premise for the thing that they’re
used for in this case (integration). He’s probably the closest related individual to these techniques; he
didn’t necessarily create them, but he invented calculus, which is what they’re used for (in this class, at
least). As I said before, it’s not really fair to try to attribute the creation of these two techniques to any
single individual because I’m sure that mathematicians for countless generations have been splitting up
fractions and using a ‘reverse-product rule’ for integration. So for me, I find it best to just credit Isaac
Newton the most because he’s the guy who’s influence most extended specifically to these two
techniques.
Exercises
[No Calculator]
Integration by Parts and Partial Fractions
Evaluate the following integrals using the integration by parts method. Show all steps and substitutions.
1.  ln xdx
2.  x sin xdx
3.  xe x dx
4.
 x
x3
2
2

2
dx
5.  x 2 x dx
Evaluate the following integrals using the integration by partial fractions method. Show all work.
 5 x  11
dx
2
 x2
6.
x
7.
 2x
8.
x
2
9.
x
2 x
dx
2
 5x
10.
3
1
dx
 x2  x
1
dx
4
x2 1
 x 2  16 dx
Implicit Differentiation
Directions: Solve for
dy
using implicit differentiation.
dx
1.
y2  7
2.
y3  x
3.
y 5  2x  3
4.
3y 2  2x  0
5.
6 y 5  y 3  y  2x 2  7  0
6.
4y3  y2  x2
7.
y 7  x 3  6x  0
8.
y 2 6 y  3x 2  6
9.
y 2 x 3  y 4  2x
y 3 6 y  7 xy
 y3
10.
2
6x
Multiple Choice
[No Calculator]
Integration by Parts and Partial Fractions
1.
 x sin( 6 x)dx 
(A)  x cos( 6 x)  sin( 6 x)  C
(C) 
x
1
cos(6 x)  sin( 6 x)  C
6
6
(B) 
(D)
x
1
cos(6 x)  sin( 6 x)  C
6
36
x
1
cos(6 x)  sin( 6 x)  C
6
36
(E) 6 x cos(6 x)  sin( 6 x)  C
2.
2x
 x  2x  1 dx 
(A) ln x  2  ln x  1  C
(B) ln x  2  ln x  1  3x  C
(C)  4 ln x  2  2 ln x  1  C
(D) 4 ln x  2  2 ln x  1  C
(E) 2 ln x 
2
1
x  x2  C
3
2
3. If f ' ( x ) 
1
then f (x) 
x 6 x  8
2
(A)
1 x4
ln
C
2 x2
(B) ln
(C)
1
ln  x  2  x  4   C
2
(D)
(E) ln x  2x  4  C
1
2
x2
C
x4
1
ln  x  2x  4  C
2
Implicit Differentiation
1. The slope of the line tangent to the curve y 2  ( xy  1) 3  0 at (2, -1) is
3
2
3
b.) 
4
a.) 
c.) 0
3
4
3
e.)
2
d.)
2. If
d2y
dy
 1  y 2 , then

dx
dx 2
a.)  2 y
b.)  y
y
c.)
1 y2
d.) y
e.)
1
2
AP Free Response
1
1. The function f is differentiable for all real numbers. The point  3,  is on

4
the graph of y  f x , and the slope at each point x, y  on the graph is given
by
dy
 y 6  2 x  .
dx
(a) Find
d2y
1
and evaluate it at the point  3,  .
2
dx
 4
(b) Find f x  , if
dy
 e x 6  2 x  . Use integration by parts method. Show all work
dx
and substitutions.
(c) Find f x  , if
dy
1
. Use integration by partial fractions method.

dx 6  2 x 6  2 x 
Show all work.
Works Cited
"Calculus: Techniques of Integration." Integration by Parts. N.p., n.d. Web. 04 June 2013.
"College Board." AP Central. N.p., n.d. Web. 04 June 2013.
"Implicit Differentiation." Implicit Differentiation. N.p., n.d. Web. 04 June 2013.
"Integration by Parts - HMC Calculus Tutorial." Integration by Parts - HMC Calculus Tutorial.
N.p., n.d. Web. 04 June 2013.
"Partial-Fraction Decomposition: General Techniques." Partial-Fraction Decomposition: General
Techniques. N.p., n.d. Web. 04 June 2013.
"Visual Calculus - Implicit Differentiation." Visual Calculus - Implicit Differentiation. N.p., n.d.
Web. 04 June 2013.