f'(x) = [(2x5 + 2)(20x3) - (5x4)(10x4)] / (2x5 + 2)2, f'(x) = -(5x4)(2x5 + 2)-2(10x4) + (2x5 + 1)-1(20x3)

dy/dx = 2(-5x2 + 3) (-10x), dy/dx = dy/dx = -2(-5x2 + 3) (10x)

dy/dx = (5x3 + 1)(15x4) + (3x5 + 1)(15x2)

dy/dx = [(3x2 + 3)(12x2 +4x) - (4x3 +2x2)(6x)] / (3x2 + 3)2, dy/dx = -(4x3 +2x2)(3x2 + 3)-2(6x) + (3x2 + 3)-1(12x2)

y' = (-2x5)(5x4) + (x5 - 2)(-10x4), y' = -100x9 +20x4

y' = x(x2 + 1)(-1/2), y' = (1/2)(x2 + 1)(-1/2) (2x)

A derivative is also ________________

the instantaneous rate of change., the instantaneous velocity., the slope.

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Derivative Fun

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Basic Derivatives #powerrule #productrule #quotientrule #chainrule #calculus #tangentline #normalline #averagerateofchange #unsimplifed

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Questions 16

Answers

1
Find f'(x).
f'(x) = [(2x5 + 2)(20x3) - (5x4)(10x4)] / (2x5 + 2)
f'(x) = [(2x5 + 2)(20x3) - (5x4)(10x4)] / (2x5 + 2)2
f'(x) = [(5x4)(10x4) - (2x5 + 2)(20x3)] / (2x5 + 2)2
f'(x) = -(5x4)(2x5 + 2)-2(10x4) + (2x5 + 1)-1(20x3)
2
Find dy/dx.
dy/dx = 2(-5x2 + 3) (10x)
dy/dx = 2(-5x2 + 3) (-10x)
dy/dx = dy/dx = -2(-5x2 + 3) (10x)
dy/dx = 2(-5x2 + 3) + (-10x)
3
Find the instantaneous rate of change of the function at the given value.
2
1
4
3
4
Find dy/dx.
dy/dx = (5x3 + 1)(15x4)(3x5 + 1)(15x2)
dy/dx = (5x3 + 1)(15x4) - (3x5 + 1)(15x2)
dy/dx = (5x3 + 1) + (3x5 + 1)
dy/dx = (5x3 + 1)(15x4) + (3x5 + 1)(15x2)
5
Uh-oh. Where is this function NOT differentiable?
At x = 1.
Nowhere! The graph is fine!
At x = 0.
At x = 2.
6
Find the equation of the TANGENT line to the function at the given point.
y = 8x +19
y = 4x +16
y = 8x - 26
y = 8x +5
7
Find the derivative at the given value.
f'(-1) = 1
f'(-1) = -1
f'(-1) = 9
f'(-1) = 6
8
Find dy/dx.
dy/dx = [(3x2 + 3)(12x2 +4x) - (4x3 +2x2)(6x)] / (3x2 + 3)2
dy/dx = -(4x3 +2x2)(3x2 + 3)-2(6x) + (3x2 + 3)-1(12x2)
dy/dx = [(3x2 + 3)(12x2 +4x) + (4x3 +2x2)(6x)] / (3x2 + 3)
dy/dx = [(4x3 +2x2)(6x) - (3x2 + 3)(12x2 +4x)] / (3x2 + 3)2
9
Find y'.
y' = 2
y' = x
y' = 2x
y' = -2
10
Find y'.
y' = (-2x5)(5x4) + (x5 - 2)(-10x4)
y' = (-2x5)(5x4) - (x5 - 2)(-10x4)
y' = (-2x5)(5x4)(x5 - 2)(-10x4)
y' = -100x9 +20x4
11
Find the average rate of change of the function over the given interval.
-3
1
-2
-1
12
Find y'.
y' = -3x3
y' = -4x4
y' = -12x3
y' = 12x3
13
The derivative is the _____________
average velocity.
instantaneous velocity.
speed.
average rate of change.
14
Find y'.
y' = (1/2)(x2 + 1)(1/2) (2x)
y' = x(x2 + 1)(-1/2)
y' = (1/2)(x2 + 1)(-1/2) (2x)
y' = (1/2)(x2 + 1)(-1/2)
15
Find y'.
y' = -3( 3x4 - 5)-4 (12x3)
y' = 3( 3x4 - 5)-4 (12x3)
y' = -3( 3x4 - 5)-2 (12x3)
y' = -3( 3x4 - 5)-2
16
A derivative is also ________________
the instantaneous rate of change.
the instantaneous velocity.
the slope.
my worst nightmare.

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