
Derivatives. Chain rule
Presentation
•
Mathematics
•
11th Grade
•
Practice Problem
•
Easy
Standards-aligned
Mariona Taniguchi
Used 15+ times
FREE Resource
4 Slides • 25 Questions
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Mastering the Chain Rule
The chain rule is a powerful tool in calculus that allows us to find the derivative of composite functions.
Helps find derivative of outer and inner functions
Product of outer function's derivative and inner function's derivative
Solves complex differentiation problems
Understanding and applying chain rule essential
Unleashes power of derivatives
3
Multiple Choice
What is the purpose of the chain rule in calculus?
To find the derivative of composite functions
To solve complex integration problems
To simplify algebraic expressions
To calculate the limit of a function
4
Chain Rule: Derivative of Composite Functions
The chain rule is a fundamental concept in calculus that allows us to find the derivative of composite functions. It is used to differentiate functions that are composed of other functions. By applying the chain rule, we can break down complex functions into simpler parts and find their derivatives. This rule is essential in various fields of mathematics and science, including physics and engineering.
Understanding composite functions and how to unleash the power of derivatives. Learn how to differentiate complex functions by breaking them down into simpler components. Apply the chain rule to find the derivative of composite functions.
5
Multiple Choice
What is the purpose of applying the chain rule to find the derivative of composite functions?
To simplify complex functions into simpler components
To understand the power of derivatives
To differentiate composite functions
To master the chain rule
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STEP 1) INDENTIFY A FUNCTION WITHIN A FUNCTION: Identify which is f(x) and which is g(x) and how they are composed f(g(x)).
STEP 2) BREAK IT DOWN: Instead of tackling it all at once, break it into parts.
First, find the derivative of the outer function f', treating g(x) as just a variable.
Then, find the derivative of the inner function g', which is the derivative of g(x).
STEP 3) MULTIPLY THEM: Now that you have the derivatives of both outer and inner functions, multiply them together, which will give the derivative of the whole composite function, f(g(x)).
Carefully watch the video and try to understand the chain rule.
Follow the steps:
7
Multiple Choice
What is the purpose of the chain rule in calculus?
To find the derivative of composite functions
To simplify complex functions
To solve equations involving derivatives
To evaluate limits of functions
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Multiple Choice
Given f(x)=2x and
g(x)=x2+3 , find (f ∘ g)(x) .
x2+2x+3
4x2+3
2x2+3
2x2+6
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Multiple Choice
Find y′ for y=(2x+1)10
10(2x+1)9
20(2x-1)9
20(2x+1)10
20(2x+1)9
10
Multiple Choice
Differentiation is also known as
Comparison
Integration
Derivatives
Multiplication
11
Multiple Choice
warning
12
Multiple Choice
What rule should be used in deriving h(x)=2(x−4)3
Chain rule
Product rule
Quotient rule
Sum rule
13
Multiple Choice
when there is a function in a function (composition of functions)
when there is division of two functions
when there is product of two functions
14
Multiple Choice
What rule will apply to find the derivative of y=xlnx
Product rule
Quotient rule
Power rule
Sum rule
15
Multiple Choice
What rule should be used in deriving f(x)=x5
Chain rule
Quotient rule
Power rule
Product rule
16
Multiple Choice
When do you use the quotient rule?
anytime you want
when there is a function in a function (composition of functions)
when there is division of two functions
when there is product of two functions
17
Multiple Choice
Find the derivative
f(x)=3x3−9x2+11x−2
f′(x)=9x2−18x+11
f′(x)=9x3−18x2+11
f′(x)=3x2−9x+11
f′(x)=9x−18
18
Multiple Choice
Use the product rule to find the derivative of
f(x)=x2sinx
f′(x)=2xcosx
f′(x)=2xsinx+x2cosx
f′(x)=2xsinx−x2cosx
f′(x)=xsinx+x2cosx
19
Multiple Choice
Use any method you want to find the derivative of
g(x)=(2x−3)(x+9)
g′(x)=2x2+15x−27
g′(x)=4x+15
g′(x)=2
g′(x)=2x
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Multiple Choice
Use division rule to find the derivative of
h(x)=xx3−9x+4
h′(x)=x2−9+4x−1
h′(x)=2x−x24
h′(x)=2x+x24
h′(x)=2x−x4
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Multiple Choice
Differentiate:
y=x32+4x41
dxdy=3x312+x431
dxdy=−3x312+x431
dxdy=3x312−x431
dxdy=−3x312−x431
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Multiple Choice
Find the derivative:
f(x)=x1−x21+x31
f′(x)=x−1−x−2+x−3
f′(x)=−x21+x32−x43
f′(x)=x21+x32−x43
f′(x)=−x21−x32−x43
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Multiple Choice
Find the derivative of f(x) = (x6 + 4)5
f '(x) = 5x5(x4 + 4)4
f '(x) = 6x5(x6 + 4)4
f '(x) = 30x5(x6 + 4)4
f '(x) = 30x6(x6 + 4)4
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Multiple Choice
Find the derivative of
f(t)=(t2+2t)5
f'(t) = 5(2t + 2)4
25
Multiple Choice
Given f(x)=(2x+1x)3 , find f′(x) .
3(2x+1x)2
3[(2x+1)21]2
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(2x+1)43x2
26
Multiple Choice
Differentiate f(x)=x2−1 .
f′(x)=x2−1
f′(x)=2xx2−11
f′(x)=x2−1x
f′(x)=xx2−1
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Multiple Choice
Differentiate f(x)=x2−31 .
f′(x)=2x−3
f′(x)=2x−31
f′(x)=(x2−3)2−2x
f′(x)=(x2−3)22x
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Multiple Choice
Differentiate f(x)=ln(x2−3) .
f′(x)=2x1
f′(x)=2x−32x
f′(x)=x2−32x
f′(x)=(x2−3)22x
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Multiple Choice
Differentiate f(x)=2x2−3 .
f′(x)=2x2−3 ⋅ 2x
f′(x)=2x2−3 ⋅ ln2 ⋅ 2x
f′(x)=2x2−3
f′(x)=2x2−3 ⋅ ln2
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