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

y' = (-3x4 - 5)(-20x3) + (-5x4 + 1)(-12x3), y' = 120x7 - 36x2 +100

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

f'(x) = (5x4 + 1)(-15x4) + (-3x5 - 4)(20x3)

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

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

y' = (x5 - 4x4 - 1)(-10x4) + (-2x5 +1)(5x4 - 16x3)

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Basic Derivatives Review

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

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Questions 19

Answers

1
Find dy/dx.
dy/dx = (5x3 + 1)(15x4) + (3x5 + 1)(15x2)
dy/dx = (5x3 + 1)(15x4) - (3x5 + 1)(15x2)
dy/dx = (5x3 + 1)(15x4)(3x5 + 1)(15x2)
dy/dx = (5x3 + 1) + (3x5 + 1)
2
Find y'
y' = (-3x4 - 5)(-20x3) + (-5x4 + 1)(-12x3)
y' = (-3x4 - 5)(-20x3) - (-5x4 + 1)(-12x3)
y' = (-3x4 - 5)+ (-5x4 + 1)
y' = 120x7 - 36x2 +100
3
Uh-oh. Where is this function NOT differentiable?
At x = 0.
At x = 1.
At x = 2.
Nowhere! The graph is fine!
4
Find dy/dx.
dy/dx = [(3x2 + 3)(12x2 +4x) + (4x3 +2x2)(6x)] / (3x2 + 3)
dy/dx = -(4x3 +2x2)(3x2 + 3)-2(6x) + (3x2 + 3)-1(12x2)
dy/dx = [(4x3 +2x2)(6x) - (3x2 + 3)(12x2 +4x)] / (3x2 + 3)2
dy/dx = [(3x2 + 3)(12x2 +4x) - (4x3 +2x2)(6x)] / (3x2 + 3)2
5
Find y'.
y' = -4x4
y' = -3x3
y' = 12x3
y' = -12x3
6
Find f'(x).
f'(x) = (-15x4) + (-3x5 - 4)
f'(x) = (5x4 + 1)(-15x4) - (-3x5 - 4)(20x3
f'(x) = (5x4 + 1)+ (-3x5 - 4)
f'(x) = (5x4 + 1)(-15x4) + (-3x5 - 4)(20x3)
7
Find the average rate of change of the function over the given interval.
-2
1
-3
-1
8
Find y'.
y' = (-2x5)(5x4)(x5 - 2)(-10x4)
y' = (-2x5)(5x4) - (x5 - 2)(-10x4)
y' = -100x9 +20x4
y' = (-2x5)(5x4) + (x5 - 2)(-10x4)
9
Find the equation of the TANGENT line to the function at the given point.
y = 8x +5
y = 8x - 26
y = 8x +19
y = 4x +16
10
Find the derivative at the given value.
f'(-1) = -1
f'(-1) = 1
f'(-1) = 6
f'(-1) = 9
11
Find f'(x).
f'(x) = [(2x5 + 2)(20x3) - (5x4)(10x4)] / (2x5 + 2)2
f'(x) = [(2x5 + 2)(20x3) - (5x4)(10x4)] / (2x5 + 2)
f'(x) = -(5x4)(2x5 + 2)-2(10x4) + (2x5 + 1)-1(20x3)
f'(x) = [(5x4)(10x4) - (2x5 + 2)(20x3)] / (2x5 + 2)2
12
Find y'.
y' = 2x
y' = x
y' = 2
y' = -2
13
Find the derivative at the given value.
f' (3) = 5
f' (3) = 5/2
f'(3) = 3
f' (3) = 13/2
14
Find y'.
y' = (x5 - 4x4 - 1)(-10x4) + (-2x5 +1)(5x4 - 16x3)
y' = (x5 - 4x4 - 1)(-10x4) - (-2x5 +1)(5x4 - 16x3)
y' = (x5 - 4x4 - 1)(-10x4)(-2x5 +1)(5x4 - 16x3)
y' = (-10x4) + (5x4 - 16x3)
15
Find the equation of the TANGENT line to the function at the given point.
y = 5x -1
y = 5x - 6
y = 3x2-2x
y = 5x + 29
16
Find the average rate of change of the function over the given interval.
-3
-4
-2
-5
17
Find the instantaneous rate of change of the function at the given value.
2
5
4
6
18
Find the derivative at the given value.
f'(3) = 45
f'(3) = 4
f'(3) = 3
f'(3) = -3
19
Find the instantaneous rate of change of the function at the given value.
2
1
4
3

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