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Derivatives with Power Rule
Derivatives with Power Rule
25 terms
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StudBud Administration
Basic Derivatives #powerrule #calculus #tangentline #averagerateofchange #unsimplifed
Flashcards
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Questions
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Answers
1
Find y'.
y' = (x5 - 4x4 - 1)(-10x4) - (-2x5 +1)(5x4 - 16x3)
y' = (-10x4) + (5x4 - 16x3)
y' = (x5 - 4x4 - 1)(-10x4) + (-2x5 +1)(5x4 - 16x3)
y' = (x5 - 4x4 - 1)(-10x4)(-2x5 +1)(5x4 - 16x3)
2
Find the derivative at the given value.
f' (3) = 5
f'(3) = 3
f' (3) = 13/2
f' (3) = 5/2
3
Find the equation of the NORMAL line to the function at the given point.
y = x -1
y = x +1
y = -x +1
y = -x +3
4
Find the instantaneous rate of change of the function at the given value.
6
4
5
2
5
Find y'.
y' = -3( 3x4 - 5)-4 (12x3)
y' = -3( 3x4 - 5)-2 (12x3)
y' = 3( 3x4 - 5)-4 (12x3)
y' = -3( 3x4 - 5)-2
6
Find dy/dx.
dy/dx = 4(3x4 + 4)3 (12x3)
dy/dx = 4(3x4 + 4)3
dy/dx = 4(3x4 + 4)3 + 12x3
dy/dx = 4(12x3)3
7
Find f'(x).
f'(x) = [(2x5 + 2)(20x3) - (5x4)(10x4)] / (2x5 + 2)2
f'(x) = [(5x4)(10x4) - (2x5 + 2)(20x3)] / (2x5 + 2)2
f'(x) = [(2x5 + 2)(20x3) - (5x4)(10x4)] / (2x5 + 2)
f'(x) = -(5x4)(2x5 + 2)-2(10x4) + (2x5 + 1)-1(20x3)
8
A derivative is also ________________
the instantaneous velocity.
my worst nightmare.
the instantaneous rate of change.
the slope.
9
Find y'.
y' = -4x4
y' = -3x3
y' = 12x3
y' = -12x3
10
Find the instantaneous rate of change of the function at the given value.
1
3
4
2
11
The original function, for example s(t), is the ____________
derivative.
time.
position function.
instantaneous rate of change.
12
Find dy/dx.
dy/dx = [(4x3 +2x2)(6x) - (3x2 + 3)(12x2 +4x)] / (3x2 + 3)2
dy/dx = [(3x2 + 3)(12x2 +4x) - (4x3 +2x2)(6x)] / (3x2 + 3)2
dy/dx = [(3x2 + 3)(12x2 +4x) + (4x3 +2x2)(6x)] / (3x2 + 3)
dy/dx = -(4x3 +2x2)(3x2 + 3)-2(6x) + (3x2 + 3)-1(12x2)
13
Find the derivative at the given value.
f'(-1) = 9
f'(-1) = -1
f'(-1) = 6
f'(-1) = 1
14
Find the equation of the NORMAL line to the function at the given point.
y = (-1/4)x -(1/2)
y = (-1/4)x +(1/2)
y = 4x +9
y = 4x - 6
15
Find y'.
y' = -2
y' = x
y' = 2
y' = 2x
16
Find the average rate of change of the function over the given interval.
3
-1
-2
-4
17
Find f'(x).
f'(x) = (5x4 + 1)+ (-3x5 - 4)
f'(x) = (5x4 + 1)(-15x4) + (-3x5 - 4)(20x3)
f'(x) = (5x4 + 1)(-15x4) - (-3x5 - 4)(20x3
f'(x) = (-15x4) + (-3x5 - 4)
18
Find the slope of the TANGENT line to the function at the given point.
4
-8
8
20
19
Find y'.
y' = (1/2)(x2 + 1)(-1/2) (2x)
y' = (1/2)(x2 + 1)(-1/2)
y' = x(x2 + 1)(-1/2)
y' = (1/2)(x2 + 1)(1/2) (2x)
20
Find the equation of the TANGENT line to the function at the given point.
y = 5x + 29
y = 5x -1
y = 3x2-2x
y = 5x - 6
21
Find the derivative at the given value.
f'(3) = -3
f'(3) = 45
f'(3) = 4
f'(3) = 3
22
Find dy/dx.
dy/dx = (5x3 + 1) + (3x5 + 1)
dy/dx = (5x3 + 1)(15x4) + (3x5 + 1)(15x2)
dy/dx = (5x3 + 1)(15x4)(3x5 + 1)(15x2)
dy/dx = (5x3 + 1)(15x4) - (3x5 + 1)(15x2)
23
Find the equation of the NORMAL line to the function at the given point.
y = -6x +14
y = -6x
y = (1/6)x +(7/3)
y = (1/6)x + (5/3)
24
Find the average rate of change of the function over the given interval.
-2
-3
-5
-4
25
Find the average rate of change of the function over the given interval.
-1
-3
-2
1
