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MATH 1271
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Scot Adams
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Midterm

Department

Mathematics

Course Code

MATH 1271

Professor

Scot Adams

Description

Priyanka Ketkar
Exam Analysis: Calculus I, Midterm 2
Test Breakdown
The main topics covered in Midterm 2 were differentiation, limits, function analysis, and
functions.
The test is usually 15-16 questions.
Part I: There are 6 multiple choice questions and the majority of those deal with differentiation,
although other topics may show up on it as well. The differentiation questions range from taking
the derivative of a complex function to merely identifying a property.
Part II: There are 5 true or false questions. The majority of these questions deal with either limits
or function analysis. There may occasionally be questions about functions in here as well. These
questions are usually straightforward and directly refer to certain rules and properties.
Part III: There are 4 or 5 free response questions that may contain multiple parts. They contain
differentiation, limits, and function analysis questions. This section requires deeper
understanding of the topic and possibly some longer hand calculations.
Test Statistics
Topics Spring 2011 Fall 2011 Spring 2012 Fall 2012 Spring 2013 Total
Differentiation 9 7 8 8 10 42
Limits 2 1 2 3 6 14
Function 5 7 6 5 2 25
Analysis
Functions 3 1 1 5 Topic 1:Differentiation
1. Knowledge Summary
Derivative: slope of a tangent to a curve; instantaneous rate of change; use tangent
formula, but there are different ways to express derivatives
o dy/dx: derivative of y with respect to x
o y’: first derivative of y; also expressed as f’(x)
o df/dx: derivative of a function with respect to x
o d/dx f(x): derivative of function f(x).
Derivatives of polynomials are always one degree less than the original function.
When comparing graphs of functions and their derivatives, maximums and minimums
occur where the slope equals the tangent.
Rules
Derivative of a Constant function: f’(c) = 0
Power Rule: if f(x) = x , then f’(x) = nx – 1
o If more than one term in a function, do this to each individual term.
Constant Multiple Rule: f’(cu) = cf’(x)
Sum and Difference Rule: f’(u ± v) = f’(u) ± f’(v)
Product Rule: f’(uv) = uf’(v) + vf’(u)
o For three functions: f”(uvw) = uvw’ + uv’w + u’vw
2
Quotient Rule: f’(u/v) = (vf’(u) – uf’(v))/v Chain Rule: if y = f(g(x)), then y’ = f’(g(x)) * g’(x); The derivative of the outside times
the derivative of the inside.
o If there are three functions when the main function is decomposed, do y’ of the
outside * y’ of the second outside * y’ of the inside
Derivatives of Exponential and Logarithmic Functions
L'Hôpital's rule: if a limit is indeterminant, take the derivative of the numerator and the
denominator separately and then put them back together. (Use only for the limit
definition, not as a substitution for the quotient rule). Only works if it is an indeterminant.
x x u u
dy/dx e = e dy/dx e = e * du/dx
a = e lna^x= exlna…. Use this to change terms so you can use exponential derivative rules
dy/dx a = a * du/dx * lna
dy/dx lnu = 1/u *(du/dx)
dy/dx log b = 1/(ulnb) * du/dx
Derivatives of Trigonometric and Inverse Trigonometric Functions
f’sinx = cosx
f’cosx = –sinx
2
f’tanx = sec x
f’cscx = –(cscx)(cotx)
f’secx = (secx)(tanx)
2
f’cotx = –csc x
f’sin (u) = 1/(1 – u ) (du/dx)
-1 2 1/2
f’cos (u) = –1/(1 – u ) (du/dx) -1 2
f’tan (u) = 1/(1 + u ) (du/dx)
f’cot (u) = –1/(1 + u ) (du/dx)
f’sec (u) = 1/(|u|(u – 1) ) (du/dx)
-1 2 1/2
f’csc (u) = –1/(|u|(u – 1) ) (du/dx)
Implicit vs. Explicit Differentiation
Explicit Function: in terms of a single variable; y equals some expression with only x and
2 2x
constants; ex: y = 2x + 3x, y = cos(╥ )
o For derivatives, just do dy/dx = and just use previous rules to manipulate x
Implicit Function: the x’s and y’s are together; may or may not be able to use algebra to
make it explicit; ex: y = sin(xy), x + 2xy + y = 14
o To take derivative, take derivative of all the terms on both sides using previous
rules
o If taking derivative of y term, multiply that term by dy/dx; for x terms, multiply
by dx/dx (this cancels because it equals 1). Basically take derivative of everything
with respect to x.
o Use algebra to solve for dy/dx
Differentiability
A function is differentiable at a certain point if the limit of its derivative exists at that
point. Thus all properties (sum, difference, product, quotient, and exponent) of limits
apply to test differentiability.
Functions are not differentiable at
o Corners: usually in piecewise functions; if multiple tangents are possible,
derivative does not exist. o Cusp (an extreme, curvy corner)
o Vertical Tangent: if the slope is going to be vertical, or undefined, derivative does
not exist
o Any type of discontinuity in the function.
*Derivatives exist at endpoints because the limit of the function at that point exists.
Not all continuous functions necessarily have continuous derivatives, but all continuous
derivatives come from continuous functions.
2. Summary of Questions to be Asked
These questions predominantly appear in Part I and Part III. The questions may simply
ask to calculate the derivative of a certain function, or to apply the derivative to find the
slope of the tangent line. Usually, the questions in Part III require the application of more
rules at one time and longer calculations than do those of Part I.
3. Related Past Test Questions
Spring 2011 Midterm 2 Part I Question A
Answer: (d) ( ) ( )( ) ( ) ( )
This question involves using the product rule f’(uv) = uf’(v) + vf’(u) and then the chain
rule (if y = f(g(x)), then y’ = f’(g(x)) * g’(x)) to take the derivative of the given
expression.
[( ) ( )] (( ) ( ))
( ) ( ) ( ) (( ) )
( )
( ) ( ) ( ) ( )( ) Spring 2011 Midterm 2 Part I Question B
( )( ) ( )( )
Answer: (a) ( )
This question requires us to know the quotient rule (f’(u/v) = (vf’(u) – uf’(v))/v ) of
x x
differentiation as well as the fact that the derivative of e is e .
( ) ( ) ( ) ( )
[ ]
( )
( )( ) ( )( )
( )
Spring 2011 Midterm 2 Part I Question C
Answer: (c) [( ) ][( ( )) ( )( )]
In this question, there is no direct rule to differentiate a function of x raised to another
function of x. Therefore, we must manipulate the problem with implicit differentiation
even though the x and y variables are separate in this case.
( )
( ) (( ) ) ( ) ( )
( )( ) ( ( )) ( ) ( )
( ( )) ( )
[ ( )]
[( ) ][( ( )) ( )( )]
Spring 2011 Midterm 2 Part I Question D Answer: (e) NONE OF THE ABOVE
This question only requires one to know that the derivative of a constant function is 0.
cos7
When looking at the function (ln4) , we find that ln4 is a constant number and cos7 is a
constant number. Raising a constant to a constant power gives some other constant
number and the derivative of that number is 0. It is not necessary to find the actual
decimal value, only to realize that the values are constant.
Spring 2011 Midterm 2 Part I Question E
Answer: (a)
This problem requires the knowledge of how to differentiate natural log functions with
the application of the chain rule.
[ ( )] ( ) ( )
Spring 2011 Midterm 2 Part I Question F
Answer: (b) ( )
This problem requires the knowledge of how to differentiate the inverse tangent function
with the application of the chain rule.
( ) ( )
[ ( )]
( ( ) ) ( ) ( ) Spring 2011 Midterm 2 Part II Question e
Answer: False
This statement is false because it is a direct contradiction to the product rule. The product
rule states that f’(uv) = uf’(v) + vf’(u), so simply multiplying the derivatives of each
individual function is the incorrect way of finding the derivative of the product of the
functions.
Spring 2011 Midterm 2 Part III Question 2
Answer: ( )
Finding the tangent line requires finding the slope and an x and y coordinate. We already
have the coordinates, so we need to find the slope, which is the same as a derivative at a
certain point. This particular equation will involve implicit differentiation.
( ) ( )
( ) ( )
( )
Now that we have the general derivative, we must evaluate it at the given point (1,2).
( )
( )
Finally, we use the slope we just found and the given point to write the equation of the
tangent line in point-slope form. ( )
Spring 2011 Midterm 2 Part III Question 3a Answer: 13
In this problem, we must first take the derivative of f(x) and then evaluate it at x = -2.
( )
( ) ( ) ( )
Fall 2011 Midterm 2 Part I Question B
Answer: (b)
Finding the logarithmic derivative is the same thing as assuming that the given function is
enclosed in a natural log function and then to take the derivative of that. For this, we need
to know how to differentiate a natural logarithmic function and apply the chain rule.
( ) ( ) ( )
Fall 2011 Midterm 2 Part I Question C
Answer: (d) 8
Finding the slope is the same as finding the derivative and evaluating it at a certain point.
First we need to find the derivative of the given function using the product rule and
knowing how to differentiate exponential functions of base e.
( )
Now, we evaluate it for the point (0,4).
( ) ( ) (( ) ) ( )
Fall 2011 Midterm 2 Part I Question D Answer: (b) ( ( )) ( )
Finding the logarithmic derivative is the same thing as assuming that the given function is
enclosed in a natural log function and then to take the derivative of that. For this, we need
to know how to differentiate a natural logarithmic function and apply the product and
chain rule.
(( ) ) ( ( ))
( ) ( ) ( ) ( )
Fall 2011 Midterm 2 Part I Question E
Answer: (a) [( ) ][( ( )) ( )]
In this question, there is no direct rule to differentiate a function of x raised to another
function of x. Therefore, we must manipulate the problem with implicit differentiation
even though the x and y variables are separate in this case.
( )
( ) (( ) ) ( )
( )( ) ( ( )) ( )
*( ( )) ( )+
[( ) ] ( ( ))
[ ( )]
Fall 2011 Midterm 2 Part III Question 1 ( ( ) ( )( )
Answer: ( )
( )
Here, we must take the derivative of the given function and apply several different rules
of differentiation.
* +
( )( ) ( )( )
( )
( )
Fall 2011 Midterm 2 Part III Question 2
( )
Answer: ( ) ( )
Here, we have no other choice than to use implicit differentiation, and that is what the
question asks us to do.
( )
(( ) ) ( ) ( )
( )( ) ( ) ( ) ( ) ( ) ( )( )
( )( ( ) ( )) ( )
( )
( )
( ) ( )
Spring 2012 Midterm 2 Part I Question A
Answer: (d) ( )( ( )) ( )( ) Here, we must assume that the given function is inside a natural logarithmic function. We
also have to remember our product rule here.
[ (( ) )] [( ) ( )]
( )( ( )) ( )( ) ( )
( )( ( )) ( )( )
Spring 2012 Midterm 2 Part I Question B
Answer: (d) [( ) ]*( )( ( )) ( ) ( )+
In this question, there is no direct rule to differentiate a function of x raised to another
function of x. Therefore, we must manipulate the problem with implicit differentiation
even though the x and y variables are separate in this case.
( )
( ) (( ) ) ( ) ( )
( )( ) ( ( )) ( )
*( )( ( )) ( ) ( )+
[( ) ]*( )( ( )) ( )( )+
Spring 2012 Midterm 2 Part I Question D
Answer: (a) y – 1 = 4(x – 1) First, we need to find the derivative of the function, which we must do here by implicit
differentiation. Then we evaluate it at the point (1,1). Finally, we can use the point and
slope (the evaluated derivative) to get the equation of the tangent line in point slope form.
( ( )) ( )
( )( )
( )
At the point (1,1):
( )
( ) ( )
The slope is 4. That along with the point (1,1) gives us the tangent line equation y – 1 =
4(x – 1), which is answer choice a.
Spring 2012 Midterm 2 Part I Question E
Answer: (e) NONE OF THE ABOVE
Differentiation of this function requires us to use the chain rule many times.
[ ( ( ))]
( ( )) ( )
This does not match with any of the answer choices, so the answer is (e) NONE OF THE
ABOVE.
Spring 2012 Midterm 2 Part I Question F Answer: (c)
Here, we must assume that the given function is inside a natural logarithmic function.
[ ( )]
Spring 2012 Midterm 2 Part III Question 1a
( ( ))( ) ( )( )
Answer: ( ( ))
Here, we must take the derivative of the given function and apply several different rules
of differentiation.
* +
( )
( ( ))( ) ( )( ) ( )
( )
( ( ))
( ( ))( ) ( )( )
( ( ))
Spring 2012 Midterm 2 Part III Question 1b
Answer: [( ) ]* ( ) ( )+
In this question, there is no direct rule to differentiate a function of x raised to another
function of x. Therefore, we must manipulate the problem with implicit differentiation
even though the x and y variables are separate in this case.
( )
( ) (( ) ) ( ) ( )( ) ( ( )) ( )
* ( ) ( )+
[( ) ] * ( ) ( )+
Spring 2012 Midterm 2 Part III Question 1b
Answer: ( )
( )
The question directly asks us to implicitly differentiate the given function.
( ) ( ( ))
( ) ( ) ( )
( ) ( ) ( )
( )
( )
Fall 2012 Midterm 2 Part I Question A
Answer: (d) ( )( ( ( ))) ( ) ( ( ))
( )
Here, we must assume that the given function is inside a natural logarithmic function. We
also have to remember our product rule here.
[ (( ( )) )] [( ) ( ( ))] ( )( ( ( ))) ( ) ( ( )) ( ( ))
( )
( )( ( ( ))) ( ) ( )
( )
Fall 2012 Midterm 2 Part I Question B
Answer: (d) [( ( )) ]*( )( ( ( ))) ( )( ( ) )+
( )
In this question, there is no direct rule to differentiate a function of x raised to another
function of x. Therefore, we must manipulate the problem with implicit differentiation
even though the x and y variables are separate in this case.
( ( ))
( ) (( ( )) ) ( ) ( ( ))
( )
( )( ) ( ( ( ))) ( )
( )
( )
[( )( ( ( ))) ( )( )]
( )
( )
[( ( )) ][( )( ( ( ))) ( )( ( ))]
Fall 2012 Midterm 2 Part I Question E
Answer: (e) NONE OF THE ABOVE
In this question, we must take the derivative with implicit differentiation.
[ ( )] ( ) ( ) ( )
This does not match any of the answer choices so the answer is (e) NONE OF THE
ABOVE Fall 2012 Midterm 2 Part I Question F
Answer: (b)
Here, we must use several laws of differentiation to take the derivative of the given
function.
[ ( )] [( ) ]
( ) ( ) ( )
Fall 2012 Midterm 2 Part II Question a
Answer: True
This question deals with differentiability and since a derivative is basically a limit, the
properties of limits apply here. Multiplying two differentiable functions at a point gives
another differentiable function at that point. Adding three differentiable functions at a
certain point gives a new differentiable function at that point. Thus, the statement is true.
Fall 2012 Midterm 2 Part III Question 1
( ( ))( ) ( )( ( ) ( ))( )
Answer: ( ( ))
We must apply several rules of differentiation here.
* +
( )
( ( ))( ) ( )( ( ) ( ))( )
( ( )) Fall 2012 Midterm 2 Part III Question 2
Answer: [( ) ]*( ) ( ( )) ( )( )+
In this question, there is no direct rule to differentiate a function of x raised to another
function of x. Therefore, we must manipulate the problem with implicit differentiation
even though the x and y variables are separate in this case.
( )
( ) (( ) ) ( ) ( )
( )( ) ( ) ( ( )) ( )( )
[( )( ( )) ( )( )]
[( ) ][( )( ( )) ( )( )]
Fall 2012 Midterm 2 Part III Question 3
Answer: ( )
Finding the tangent line involves finding the slope (derivative) and then using that and
the given point to get the equation of a line in point slope form.
( )
At the point (2,1) ( ) ( )
( )
With this slope and the point (2,1), we get the tangent line to be ( )
Spring 2013 Midterm 2 Part I Question A
Answer: (a)
Here, we are given a point and the first derivative. We must evaluate the derivative at the
point when x = 0. f’(0) = 4. With this slope and the given point, we find the equation of
the tangent line to be y – 1 = 4(x – 0) y = 1 + 4x. This matches choice a.
Spring 2013 Midterm 2 Part I Question C
Answer: (c) ( )
This is a direct reference to the quotient rule of differentiation and answer c correctly
expresses the quotient rule.
Spring 2013 Midterm 2 Part I Question D
Answer: (a)
Here, we can use the product property of logarithms to split the expression and then use
the sum property to take the derivative.
[ |( )( )|] [ | | | |]
Spring 2013 Midterm 2 Part II Question a
Answer: False This is a direct reference to the exponent rule of differentiation. The statement is false
because [ ] . We need to account for the chain rule.
Spring 2013 Midterm 2 Part III Question 5a
Answer: a=3, b=3, c=1
3 3
We can think of Δy as (x + Δx) – x . This is also the numerator portion of the equation
of a secant line. When fully expanded, it equals x + 3x (Δx) + 3x(Δx) + (Δx) – x . The 3
cubed terms canceled leaving us with an expression that follows the format of the one
given in the question. Thus, the coefficients are 3, 3, and 1 for a, b, and c respectively.
Spring 2013 Midterm 2 Part III Question 5b
Answer: 3x + 3x(Δx) + (Δx) 2
To do this calculation, we have to merely divide the expression we found in part a,
3x (Δx) + 3x(Δx) + (Δx) , by a Δx factor. Thus, we get 3x + 3x(Δx) + (Δx) . 2
Topic 2: Limits
1. Knowledge Summary
The limit of a function is the value the function approaches as x approaches a certain
value.
The expression ( ) refers to the limit as x approaches a from the right.
The expression ( ) refers to the limit as x approaches a from the left. The expression The expression ( ) refers to the limit as x approaches a from both
sides.
A horizontal asymptote of a function is also the limit of that function to either + or – ∞.
Properties
1) Sum and difference: The limit of a sum/difference of two functions is the
sum/difference of the limits of the separate functions( ( ) ( ))
( ) ( )
2) Product or quotient: The limit of a product/quotient of two functions is the
product/quotient of the limits of the separate functions (as long as the
( )
deno

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