MATH 165 (Calculus I) Test 2 Review Sheet - Cohen

Introduction

This review sheet covers the key topics for Test 2 in MATH 165 (Calculus I) as outlined by Professor Cohen. Test 2 focuses on differentiation techniques, applications of derivatives, and related rates problems. This material builds upon the fundamental limits and derivatives covered in Test 1.

Test Format and Weighting

• Test 2 will consist of approximately 8-10 problems
• The test is designed to be completed in 75 minutes
• No calculators or formula sheets are permitted
• All work must be shown for full credit
• Partial credit may be awarded for correct approaches even with calculation errors

Key Topics Covered

1. Differentiation Rules
2. Chain Rule and Composite Functions
3. Implicit Differentiation
4. Related Rates Problems
5. Applications of Derivatives
6. Curve Sketching Using Derivatives

Differentiation Rules Review

Master the following basic differentiation rules:

Power Rule

Constant Multiple Rule

Sum/Difference Rules

Product Rule

Quotient Rule

Chain Rule

Differentiating Specific Functions

Trigonometric Functions

• d/dx [sin(x)] = cos(x)
• d/dx [cos(x)] = -sin(x)
• d/dx [tan(x)] = sec(x)
• d/dx [sec(x)] = sec(x)tan(x)
• d/dx [csc(x)] = -csc(x)cot(x)
• d/dx [cot(x)] = -csc(x)

Exponential and Logarithmic Functions

• d/dx [e] = e
• d/dx [ln(x)] = 1/x

Chain Rule Applications

The chain rule is essential for differentiating composite functions. Remember to work from the outside in:

Implicit Differentiation

Related Rates Problems

Applications of Derivatives

Finding Maxima and Minima

To find local extrema:

1. Find critical points by setting f'(x) = 0 or where f'(x) is undefined
2. Use the First Derivative Test or Second Derivative Test to classify critical points

Absolute Extrema on Closed Intervals

To find absolute maximum and minimum on interval [a,b]:

1. Find critical numbers of f in (a,b)
2. Evaluate f at the critical numbers
3. Evaluate f at the endpoints a and b
4. The largest value is the absolute maximum; the smallest is the absolute minimum

Curve Sketching Using Derivatives

Use the following information to sketch the graph of a function:

• Domain and range
• x and y intercepts
• Symmetry (even, odd, periodic)
• Asymptotes (vertical and horizontal)
• Intervals of increase/decrease (first derivative)
• Local maxima and minima
• Intervals of concavity (second derivative)
• Points of inflection

Practice Problems

Additional Study Tips

• Practice deriving from scratch without looking at notes
• Focus on understanding conceptually rather than just memorizing
• Pay attention to domain restrictions when solving problems
• Double-check your work by plugging back into original equations
• For related rates, clearly define your variables before starting
• Try to identify patterns in similar problems
• Make sure you know the derivatives of all basic functions
• Practice working at a speed that allows you to complete problems in time

Common Mistakes to Avoid

• Forgetting the chain rule on composite functions
• Mixing up the order in the quotient rule
• Not simplifying answers completely
• Forgetting to differentiate both sides in implicit differentiation
• Not checking domain restrictions
• Confusing local and absolute extrema
• Missing critical points where a derivative is undefined
• Algebraic errors when solving equations

Formula Sheet to Remember

Derivative Rules:

• Power Rule: d/dx [x] = nx
• Product Rule: d/dx [fg] = f'g + fg'
• Quotient Rule: d/dx [f/g] = [f'g - fg']/g
• Chain Rule: d/dx [f(g)] = f'(g)g'

Common Derivatives:

• d/dx [sin(x)] = cos(x)
• d/dx [cos(x)] = -sin(x)
• d/dx [tan(x)] = sec(x)
• d/dx [e] = e
• d/dx [ln(x)] = 1/x
• d/dx [sin(x)] = 1/(1-x)
• d/dx [tan(x)] = 1/(1+x)

Test Day Preparation

• Get a good night's sleep before the test
• Arrive early with all necessary materials (ID, pen, pencil)
• Read through the entire test before beginning
• Manage your time wisely - spend appropriate time on each problem
• Show all your work clearly
• Check your answers if time permits
• If you get stuck, move on and come back later
• Breathe and stay calm throughout the test

Answers to Practice Problems

Problem 1:

Using the chain rule twice:

f'(x) = (1/cos(x))(-sin(x))2x

f'(x) = -[2xsin(x)]/cos(x)

f'(x) = -2xtan(x)

Problem 2:

Differentiating both sides with respect to x:

cos(x)y + sin(x)2ydy/dx = 3x

2ysin(x)dy/dx = 3x - cos(x)y

dy/dx = [3x - cos(x)y]/[2ysin(x)]

Problem 3:

The volume of a sphere is V = (4/3)r

Differentiating with respect to time:

dV/dt = 4rdr/dt

Given dV/dt = 10 cm/s and r = 5 cm:

10 = 4(5)dr/dt

10 = 100dr/dt

dr/dt = 10/(100) = 1/(10) 0.032 cm/s

Problem 4:

f'(x) = 4x - 12x = 4x(x - 3)

Critical points: x = 0 and x = 3

Using the second derivative test:

f''(x) = 12x - 24x

f''(0) = 12(0) - 24(0) = 0 (inconclusive)

f''(3) = 12(3) - 24(3) = 108 - 72 = 36 > 0, so x = 3 is a local minimum

Using the first derivative test for x = 0:

For x < 0: f'(x) < 0

For x > 0 but small: f'(x) < 0

The derivative does not change sign at x = 0, so there is no extremum at x = 0

The local minimum value is f(3) = (3) - 4(3) + 10 = 81 - 108 + 10 = -17

Problem 5:

f'(x) = 3x - 6x = 3x(x - 2)

Critical points in [-2,3]: x = 0 and x = 2

Evaluating f at critical points and endpoints:

f(-2) = (-2) - 3(-2) + 12 = -8 - 12 + 12 = -8

f(0) = (0) - 3(0) + 12 = 12

f(2) = (2) - 3(2) + 12 = 8 - 12 + 12 = 8

f(3) = (3) - 3(3) + 12 = 27 - 27 + 12 = 12

The absolute maximum is 12 at x = 0 and x = 3

The absolute minimum is -8 at x = -2

Problem 6:

f'(x) = 3x - 12x + 9 = 3(x - 4x + 3) = 3(x-1)(x-3)

f''(x) = 6x - 12 = 6(x-2)

Critical points: x = 1 and x = 3

Inflection point: x = 2

Intervals of increase: x < 1 and x > 3

Intervals of decrease: 1 < x < 3

Concave down: x < 2

Concave up: x > 2

Problem 7:

Given: y = x, dx/dt = 5 when x = 2

Need: dy/dt when x = 2

dy/dt = d/dt[x] = 2xdx/dt

When x = 2, dy/dt = 2(2)(5) = 20

Therefore, the y-coordinate is changing at a rate of 20 units per second

Problem 8:

Using implicit differentiation:

2x + 2(xdy/dx + y) + 2ydy/dx = 0

2x + 2xdy/dx + 2y + 2ydy/dx = 0

2xdy/dx + 2ydy/dx = -2x - 2y

dy/dx(2x + 2y) = -2(x + y)

dy/dx = -2(x + y)/[2(x + y)] = -1 when x + y 0

At the point (1,1), the slope is -1

Using point-slope form: y - y = m(x - x)

y - 1 = -1(x - 1)

y - 1 = -x + 1

y = -x + 2

Therefore, the equation of the tangent line is y = -x + 2