Bell Ringer- New Website Additions
- Availability
- Post-Assessment Reflection
- Differentiation Test 2 on this Friday, 10.03.14!
- Summative Exam 1 on next Friday, 10.10.14!
Calculate dy/dx for the following: y = x dy/dx = x dy/dx = y dy/dx = 1 dy/dx = 0 none of the above
2y = x dy/dx = x/2 dy/dx = 2y dy/dx = 2 dy/dx = 0 none of the above
xy = 1 dy/dx = -x2 dy/dx = -1/x2 dy/dx = 1/x2 dy/dx = -2/x2 none of the above
x2y = x + 1 dy/dx = (-x - 2)/x3 dy/dx = (x - 2)/x3 dy/dx = (-x - 2)/x2 dy/dx = (-x - 2)/x2 none of the above
y3 + y2 - 5y - x2 = -4 dy/dx = (2x) / (3y2 + 2y - 5) dy/dx = (2x) / (3y2 + 2y + 5) dy/dx = (-2x) / (3y2 + 2y - 5) dy/dx = 2 / (3y2 + 2y - 5) - none of the above
Review
- Secant vs. Tangent Line
- Definition of Derivative
- Importance of Derivative
- What does it allow us to do?
- Drawing a Tangent Line on a Graph
- Basic Differentiation Rules
- Constant Rule
- Power Rule
- Sum and Difference Rule
- Sine and Cosine Derivatives
- Derivative Notation
- Differentiation
- Rates of Change
- Position Function
- Average Velocity
- Instantaneous Velocity
- Free Fall Problems
- Product Rule
- Quotient Rule
- Chain Rule
Lesson
- Implicit Differentiation
- Checkpoints
- A - page 146 #6
- B - page 146 #10
- C - page 146 #26
- D - page 146 #30
- E - page 147 #48
- F - page 147 #52
- G - page 147 #66
- H - page 148 #74
- I - page 148 #78
Exit Ticket
- Exit Ticket will be posted on the board in class.
| Lesson Objective(s)- How can implicit differentiation be used to find the derivative?
Standard(s) - APC.5
APC.6 APC.7 APC.8 APC.9
Mathematical Practice(s) - #1 - Make sense of problems and persevere in solving them
- #2 - Reason abstractly and quantitatively
- #3 - Construct viable arguments and critique the reasoning of others
- #4 - Model with mathematics
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