Derivative of absolute value function

Prime Newtons
4 Sept 202308:04
EducationalLearning
32 Likes 10 Comments

TLDRThis educational video script guides viewers through the process of differentiating the function y equals the absolute value of 2x minus 3. It begins by explaining why the simple differentiation of y=2x to get a derivative of 2 is not applicable here due to the absolute value function. The script then introduces the concept that the absolute value of x can be rewritten as the square root of x squared, which is a key insight for differentiation. Using the chain rule, the video demonstrates how to differentiate y=|x|, resulting in dy/dx = x/|x|. Finally, it applies a similar method to differentiate y=|2x-3|, introducing a substitution with u=2x-3 and using the chain rule to find dy/dx = (2x-3)/|2x-3|. The script emphasizes the importance of understanding the underlying concepts and encourages continuous learning.

Takeaways
  • 📚 The script discusses the differentiation of the function y = |2x - 3|, emphasizing that it's not as straightforward as differentiating a simple polynomial.
  • 🔍 It clarifies that the derivative of an absolute value function isn't simply 1 or -1, as there are two directions (positive and negative) to consider.
  • 📐 The video introduces an alternative representation of the absolute value function, suggesting that |x| can be rewritten as √(x^2), which is a key insight for differentiation.
  • 👨‍🏫 The presenter advocates for teaching the square root representation of absolute value from the beginning, as it simplifies the process of finding derivatives.
  • 🧩 The differentiation of y = |x| is demonstrated using the chain rule, showing that the derivative is x / |x|, not a constant as one might initially assume.
  • 🔄 The process involves recognizing that |x| can be expressed as x^2 raised to the power of 1/2, and then applying the chain rule to differentiate it.
  • 📉 The script uses the example of differentiating y = |2x - 3| by introducing a substitution, letting u = 2x - 3, to simplify the differentiation process.
  • 🔑 The chain rule is applied again to find the derivative of y = |u|, resulting in the expression (2u) / |u|, where u is the substituted variable.
  • 📝 The final derivative of y = |2x - 3| is given as (2(2x - 3)) / |2x - 3|, which is a product of the derivative of the inside function and the inside function itself, divided by the absolute value of the inside function.
  • 🚀 The video concludes with an encouragement to continue learning, highlighting that learning is an ongoing process essential to living.
Q & A

  • What is the main topic of the video?

    -The main topic of the video is to demonstrate the process of differentiating the function y = |2x - 3|, including the concept of absolute value and its derivative.

  • Why is the differentiation of y = |2x - 3| not straightforward?

    -The differentiation of y = |2x - 3| is not straightforward because the absolute value function can have two possible expressions, one for x ≥ 0 and another for x < 0, which means the derivative cannot be a single constant value like in simpler functions.

  • What is the initial misconception the video aims to correct regarding the absolute value function?

    -The initial misconception the video aims to correct is that the derivative of the absolute value function is not simply 1 or -1, but rather a more complex expression that depends on the variable inside the absolute value.

  • How does the video suggest rewriting the absolute value of x?

    -The video suggests rewriting the absolute value of x as the square root of x squared, which is a more mathematically rigorous representation of the absolute value function.

  • What mathematical rule is used to differentiate the absolute value of x?

    -The video uses the chain rule to differentiate the absolute value of x, by expressing it as x squared raised to the power of one half.

  • What is the derivative of y = |x| according to the video?

    -According to the video, the derivative of y = |x| is x divided by the absolute value of x, which can be written as x / |x|.

  • What substitution is made to simplify the differentiation of y = |2x - 3|?

    -The substitution made to simplify the differentiation of y = |2x - 3| is letting u = 2x - 3, which allows the use of the chain rule to find the derivative.

  • What is the derivative of u with respect to x in the context of the video?

    -In the context of the video, the derivative of u with respect to x, where u = 2x - 3, is 2.

  • How is the chain rule applied to find the derivative of y = |2x - 3|?

    -The chain rule is applied by multiplying the derivative of the outer function (dy/du, which is u / |u|) by the derivative of the inner function (du/dx, which is 2).

  • What is the final expression for the derivative of y = |2x - 3| presented in the video?

    -The final expression for the derivative of y = |2x - 3| presented in the video is 2(2x - 3) divided by the absolute value of (2x - 3).

Outlines
00:00
📚 Introduction to Differentiating Absolute Value Functions

This paragraph introduces the concept of differentiating absolute value functions, starting with a simple example of 'y = |2x - 3|'. The speaker emphasizes that the derivative of an absolute value function is not straightforward and cannot be answered with a single value. The paragraph sets the stage for a deeper exploration into the differentiation process, hinting at the complexity involved due to the nature of the absolute value function, which always results in a 'V' shape. The speaker also suggests a common misconception about the derivative of absolute value functions and hints at a more nuanced understanding that will be developed throughout the video.

05:00
🔍 Derivative of Absolute Value of x and Chain Rule Application

The speaker begins by differentiating the simpler case of 'y = |x|', explaining that the result is not a constant value but rather a function of x. They introduce the concept that the absolute value of x can be rewritten as the square root of x squared, which is a crucial insight for understanding the differentiation process. The chain rule is then applied to differentiate 'y = |x|', resulting in the derivative 'dy/dx = x / |x|'. This part of the script serves as a foundation for understanding how to differentiate more complex absolute value functions, such as 'y = |2x - 3|', which will be tackled in the subsequent paragraph.

📘 Differentiating Absolute Value of 2x - 3 Using U-Substitution

Building upon the previous discussion, the speaker now addresses the differentiation of 'y = |2x - 3|'. They introduce a U-substitution technique where U is set to '2x - 3', simplifying the problem to differentiating 'y = |U|'. The derivative of U with respect to x (du/dx) is identified as 2. The chain rule is then applied again, leading to the final derivative 'dy/dx = (2 * (2x - 3)) / |2x - 3|'. This step-by-step process illustrates the application of the chain rule and U-substitution in the context of absolute value functions, providing a clear path to the solution and reinforcing the learning from the earlier examples.

Mindmap
Keywords
💡Differentiation
Differentiation in the context of the video refers to the process of finding the derivative of a mathematical function. It is a fundamental concept in calculus and is used to determine the rate at which a function changes. The video's theme revolves around teaching the differentiation of functions, specifically those involving absolute values, which are more complex than basic differentiation.
💡Absolute Value
The absolute value of a number is the non-negative value of that number without regard to its sign. In the video, absolute value is used in the function y = |2x - 3|, and understanding its properties is crucial for differentiating the function correctly. The script explains that the absolute value function can be rewritten as the square root of the number squared, which is a key insight for the differentiation process.
💡Derivative
A derivative in calculus represents the rate at which a function changes with respect to its variable. The video script discusses finding the derivative of the absolute value function, which is not straightforward and requires the use of the chain rule and understanding the behavior of the function inside the absolute value.
💡Chain Rule
The chain rule is a fundamental theorem in calculus for differentiating composite functions. It states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. In the video, the chain rule is applied to differentiate y = |x| and y = |2x - 3|, where the outer function is the square root and the inner function is x squared or 2x - 3, respectively.
💡Square Root
A square root is a value that, when multiplied by itself, gives the original number. The video script mentions that the square root of x squared, denoted as √(x^2), is equivalent to the absolute value of x. This equivalence is used to rewrite the absolute value function in a form that can be differentiated using the chain rule.
💡U-Substitution
U-substitution is a technique used in calculus to simplify the process of integration or differentiation of complex functions. In the video, the concept is analogously applied by letting u = 2x - 3 to simplify the differentiation of y = |2x - 3|. This substitution allows the use of the chain rule to find the derivative more easily.
💡Function
In mathematics, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. The video discusses differentiating various functions, particularly those involving absolute values, to understand how these functions change with their inputs.
💡Composite Functions
A composite function is a function composed of two or more functions, where the output of one function becomes the input of the next. The video script uses composite functions in the form of absolute values applied to expressions like x and 2x - 3, which are then differentiated using the chain rule.
💡Square Root Function
The square root function is a mathematical function that maps a number to its non-negative root. In the video, the square root function is used in the context of differentiating the absolute value function, as the square root of x squared (√(x^2)) is shown to be equivalent to the absolute value of x.
💡Learning
The concept of learning is emphasized in the video as an ongoing process. The script mentions that there is no end to learning, and by extension, no end to living. This philosophical note underscores the importance of continuous education and the pursuit of knowledge, which is particularly relevant in the context of mastering complex mathematical concepts like differentiation.
Highlights

Differentiating the absolute value function requires understanding its behavior differently from simple derivatives.

The derivative of a simple expression like y=2x would be 2, but the absolute value changes this.

The absolute value function is always non-negative, which complicates its derivative.

The absolute value of x can be rewritten as the square root of x squared.

Differentiating the absolute value of x is not straightforward and involves the chain rule.

The chain rule is applied to differentiate y = |x| by considering it as (x^2)^(1/2).

The derivative of y = |x| is x divided by the absolute value of x.

Differentiating y = |2x - 3| involves a U-substitution with U = 2x - 3.

The derivative of U = 2x - 3 with respect to x is 2.

The chain rule is used again to find the derivative of y = |U| with respect to x.

The derivative of y = |U| with respect to x is 2U/|U| where U = 2x - 3.

The final derivative of y = |2x - 3| is (2(2x - 3))/(|2x - 3|).

Differentiating absolute value functions requires understanding their piecewise nature.

The process of differentiating absolute value functions is a good example of applying the chain rule and understanding piecewise functions.

Learning to differentiate absolute value functions is a key skill in calculus.

Transcripts

Browse More Related Video

Implicit Differentiation

Implicit Differentiation Method

Implicit Differentiation for Calculus - More Examples, #1

Average Value of a Function

Finding particular linear solution to differential equation | Khan Academy

Absolute Maximum and Minimum Values - Finding absolute MAX & MIN of Functions - Calculus

Rate This

5.0 / 5 (0 votes)

Thanks for rating:

Related Tags
CalculusDifferentiationAbsolute ValueChain RuleDerivativeMathematicsEducationalLearningMath TutorialComplex Functions