Download Differentiation Exam Review

DIFFERENTIATION
TEST REVIEW
AP CALCULUS
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HEY, PEOPLE--WHAT ARE THE THREE WAYS A
FUNCTION CAN BE NON-DIFFERENTIABLE?
1) NON-CONTINUOUS
2) VERTICAL TANGENT LINE
3) SHARP CORNER/TURN
PLEASE FIND F’(X)
F(X)=4SIN(2X)
F’(X)=8cos(2X)
READY…SET….GO!
Find the f’(4)
2x + 4
F(x) =
x-6
f’(x) = -16 / (x-6)^2
f’(4) = -16/ (4-6) ^2
f’(4) = -4
PLEASE FIND F’(X)
X +3
F(X) =
X -2
F’(x) = -5 / (x-2)^2
BONUS!
If f(x) = cosx, then f’(pi/6) =
F’(x)=-sinx
f’(pi/6) = -sin(pi/6) = -0.5
Find f’(4)
16
f (x) = 8 x + + 3
x
F’(x) = 1
If
dy/dx = -12sin6xcos6x
If f(x) = (3x-2)^3, then f”(1) =
F’(x) = 9(3x-2)^2
F”(x) = 54(3x-2)
F”(1) = 54(3-2) = 54
If dy/dx=
Then y”=
y” = 1 (WHAT?!)
2y + 6
BONUS!
If f(x) = 4cos(x/2), then f”(pi/2) =
f‘(x) = -2sin(x/2)
f”(x) = -cos(x/2)
f”(pi/2) = -cos(pi/4)
f”(pi/2) = - (sqrt2) / 2
If y= tan5x, then y’ =
y’= 5sec^2(5x)
The
=[d/dx] 8sinx
= 8 cosx
BONUS!
The
= [d/dx] 5x^2
= 10x
If f(x) = sinx * (x+8)^2, then f’(0)=
f’(x) = sinx [2(x+8)] + cosx[(x+8)^2]
f’(0) = 0(16) + (1)(64)
F’(0) = 64
If f(x) = 2x^(5/2), then f’(9)=
f’(x) = 5x^3/2
f’(9) = 5* 9^3/2
f’(9) = 5 * 27
f’(9) = 135
If
f (x) = x 4x +1
Then f’(2) =
f’(x) =x * 2/sqrt(4x+1) + sqrt(4x+1)
f’(x) = [6x+1] / sqrt(4x+1)
f’(x) = 13/3
READY…SET….GO!
What is the instantaneous rate of change at
x=3?
3+ x ^ 2
2x -1
f’(x) = [(2x-1)(2x) – (3+x^2)(2) ]/ (2x-1)^2
f’(3) = [(5*6) – (3+9)*2] / 5^2
f’(3) = 6/25
At x=5, is f(x) continuous?
Differentiable?
YES and YES
BONUS!
At x=6, is f(x) continuous? Differentiable?
NOOO!
Remember, if it’s not continuous, it’s
most definitely not differentiable!
Is y= abs(4-2x) ever non-differentiable? Explain.
Why yes! At x=2, there is a sharp turn, so it’s not
differentiable there.
BONUS!
2x + xy + y^2 = 3x -5
dy/dx =?
dy/dx = [1-y] / [x+2y]
4y + 6x = 2y^2
dy/dx at (1,3):
4y’ + 6 = 4yy’
4y’ – 4yy’ = -6
y’(4-4y) = -6
y’= -6 / (4-4y)…. = -6/(4-12) = 3/4
READY…SET….GO!
s(t) = 3t^3 – 18t^2 -45t
What is the velocity function?
v(t) = s’(t) = 9t^2 – 36t – 45
s(t) = 3t^3 – 18t^2 -45t
When is the velocity zero?
v(t) = 9t^2 – 36t -45 = 0
9 (t^2 -4t – 5)= 0
9(t-5)(t+1) = 0
t=5, -1
t=5 seconds
BONUS!
s(t) = 3t^3 – 18t^2 -45t
What is the acceleration function?
v(t) = 9t^2 – 36t – 45
a(t) = v’(t) = 18t - 36
s(t) = 3t^3 – 18t^2 -45t
When is the acceleration zero?
a(t) = v’(t) = 18t – 36 = 0
18t = 36
t=2 seconds
Let’s finish on an easy note!
f(x) = 2x
f’(x)=
f’(x) = 2
Good luck on your exam!
Be sure to take time
and study!
The test WILL NOT be easy, but you can and will do
well if you put in the time and effort!
*All topics we’ve covered are fair game, and listed under the
notes tab of the course website
*Past homework assignments should be your main studying
tool. Look them over, compare with the solutions, and make
sure you understand everything.
* Remember, this test will be calculator-free. Make sure you’re
comfortable working without one, and that you can handle
both trig and fractional/negative exponents.