Derivatives - Derivatives Of Composite Functions- Chain Rule

Derivatives - Derivatives of composite functions- Chain Rule

Slide 1

The concept of derivatives is an important topic in calculus.
The derivative of a function represents its rate of change at any given point.
In this lecture, we will focus on the derivatives of composite functions using the chain rule.

Slide 2

A composite function is a combination of two or more functions.
If we have two functions, f(x) and g(x), the composite function is denoted as f(g(x)).
For example, if f(x) = x^2 and g(x) = sin(x), then the composite function is f(g(x)) = (sin(x))^2.

Slide 3

The chain rule is a method used to find the derivative of a composite function.
It states that if we have a composite function y = f(g(x)), the derivative dy/dx can be calculated using the following formula: dy/dx = (df/dg) * (dg/dx)

Slide 4

Let’s understand the chain rule with an example.
Consider the function y = (x^2 + 3x)^3.
In this case, f(u) = u^3 and g(x) = x^2 + 3x.
We need to find dy/dx.

Slide 5

First, we find the derivative of f(u), which is df/du = 3u^2.
Then, we find the derivative of g(x), which is dg/dx = 2x + 3.
Applying the chain rule, dy/dx = (df/dg) * (dg/dx) = (3u^2) * (2x + 3).

Slide 6

Now, substitute g(x) into the formula: dy/dx = (3(x^2 + 3x)^2) * (2x + 3).
Simplify the equation by expanding and combining like terms.
Finally, you will have the derivative of the composite function: dy/dx = 6x(x^2 + 3x)^2 + 9(x^2 + 3x)^2.

Slide 7

Let’s take another example to practice the chain rule.
Consider the function y = sqrt(sin(2x)).
In this case, f(u) = sqrt(u) and g(x) = sin(2x).

Slide 8

First, find the derivative of f(u), which is df/du = 1 / (2 * sqrt(u)).
Then, find the derivative of g(x), which is dg/dx = 2 * cos(2x).
Applying the chain rule, dy/dx = (df/dg) * (dg/dx) = (1 / (2 * sqrt(u))) * (2 * cos(2x)).

Slide 9

Substitute g(x) into the formula: dy/dx = (1 / (2 * sqrt(sin(2x)))) * (2 * cos(2x)).
Simplify the equation by canceling out the common factors.
Finally, you will have the derivative of the composite function: dy/dx = cos(2x) / sqrt(sin(2x)).

Slide 10

The chain rule is a powerful tool in finding derivatives of composite functions.
It helps us break down complicated functions into simpler parts.
Practice applying the chain rule with different examples to enhance your understanding.

Slide 11

The chain rule can be applied to more complex composite functions.
It can be extended to functions with multiple layers of composition.
Let’s consider an example to illustrate this concept.

Slide 12

Suppose we have the composite function y = (sin(x^2))^3.
In this case, f(u) = u^3 and g(x) = sin(x^2).
We need to find the derivative dy/dx.

Slide 13

First, find the derivative of f(u), which is df/du = 3u^2.
Then, find the derivative of g(x) in terms of u.
Use the chain rule to find dg/dx = (dg/du) * (du/dx).

Slide 14

For g(x) = sin(x^2), dg/dx becomes (dg/du) * (du/dx).
The derivative of sin(u) is cos(u), so (dg/du) = cos(x^2).
To find (du/dx), differentiate x^2 with respect to x.

Slide 15

Differentiating x^2 with respect to x, we get (du/dx) = 2x.
Now, we have (dg/du) = cos(x^2) and (du/dx) = 2x.
Applying the chain rule, dy/dx = (df/dg) * (dg/du) * (du/dx).

Slide 16

Substitute g(x) and u into the formula: dy/dx = (3(u^3)) * (cos(x^2)) * (2x).
Simplify the equation by combining the terms.
Finally, you will have the derivative of the composite function: dy/dx = 6x(u^3)cos(x^2).

Slide 17

The chain rule can also be applied to functions with inverse trigonometric functions.
Let’s consider an example to explore this concept.

Slide 18

Suppose we have the composite function y = sec(arcsin(x)).
In this case, f(u) = sec(u) and g(x) = arcsin(x).
We want to find the derivative dy/dx.

Slide 19

First, find the derivative of f(u), which is df/du = sec(u) * tan(u).
Then, find the derivative of g(x), which is dg/dx in terms of u.
Use the chain rule to find dg/dx = (dg/du) * (du/dx).

Slide 20

For g(x) = arcsin(x), dg/dx becomes (dg/du) * (du/dx).
The derivative of arcsin(u) is 1 / sqrt(1 - u^2), so (dg/du) = 1 / sqrt(1 - x^2).
To find (du/dx), differentiate x with respect to x.

Slide 21

Differentiating x with respect to x, we get (du/dx) = 1.
Now, we have (dg/du) = 1 / sqrt(1 - x^2) and (du/dx) = 1.
Applying the chain rule, dy/dx = (df/dg) * (dg/du) * (du/dx).

Slide 22

Substitute g(x) and u into the formula: dy/dx = (sec(u) * tan(u)) * (1 / sqrt(1 - x^2)) * 1.
Simplify the equation by combining the terms.
Finally, you will have the derivative of the composite function: dy/dx = sec(arcsin(x)) * tan(arcsin(x)) * (1 / sqrt(1 - x^2)).

Slide 23

The chain rule can be applied to a wide range of composite functions.
It is a fundamental concept in calculus and is used in various applications.
Practice different examples to master the chain rule and its application in finding derivatives of composite functions.

Slide 24

Let’s summarize the key points discussed in this lecture:
- A composite function is a combination of two or more functions.
- The chain rule is used to find the derivative of a composite function.
- The formula for the chain rule is: dy/dx = (df/dg) * (dg/dx).
- The chain rule can be applied to functions with multiple layers of composition.
- It can also be used with inverse trigonometric functions.

Slide 25

Importance of understanding the chain rule:
- It helps in finding derivatives of composite functions efficiently.
- It enables us to break down complex functions into simpler parts for analysis.
- Understanding the chain rule is crucial for further studies in calculus and related topics.

Slide 26

Applications of the chain rule:
- It is used in physics to find rates of change and velocity.
- It is used in engineering to determine optimal solutions and analyze systems.
- It is used in economics and finance for modeling and optimization problems.
- It is used in computer science for image processing and machine learning algorithms.

Slide 27

Further topics to explore:
- Higher order derivatives and their applications.
- Implicit differentiation and related concepts.
- Related rates problems and their solutions using the chain rule.
- Optimization problems and finding local/global extrema using derivatives.

Slide 28

Recap of the main takeaways from this lecture:
- The chain rule is used to find derivatives of composite functions.
- It involves finding the derivatives of individual functions and combining them using the chain rule formula.
- The chain rule can be applied to functions with multiple layers of composition.
- It is a fundamental concept in calculus and has various applications in different fields.

Slide 29

Resources for further study:
- Textbooks: Stewart’s Calculus, Thomas’ Calculus, and Apostol’s Calculus.
- Online resources: Khan Academy, MIT OpenCourseWare, and Coursera courses on calculus.
- Practice problems and exercises available in textbooks and online platforms.

Slide 30

Conclusion:
- The chain rule is an essential tool in calculus for finding derivatives of composite functions.
- It allows us to analyze complex functions by breaking them down into simpler parts.
- Understanding and applying the chain rule is crucial for tackling advanced calculus problems and real-world applications.