Finding derivative with fundamental theorem of calculus: chain rule | AP®︎ Calculus | Khan Academy

Khan Academy
7 Aug 201903:07
EducationalLearning
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TLDRThe video script presents a mathematical problem involving the calculation of the derivative of a function F(x), which is defined as a definite integral with an upper bound of sine(x). The challenge lies in the fact that the upper bound is not a constant x, but a function of x, sine(x). To find F'(x), the instructor suggests using the chain rule by first defining g(x) as sine(x), which allows the function F(x) to be expressed as h(g(x)). The derivative F'(x) is then found by multiplying the derivative of h with respect to g(x), h'(g(x)), with the derivative of g(x), g'(x). The inner function h'(x) is identified as 2x - 1, and since g(x) is sine(x), g'(x) is the derivative of sine, which is cosine(x). Thus, F'(x) is the product of 2 times sine of x and cosine of x, providing a clear solution to the problem.

Takeaways
  • 📚 The function F(x) is defined as a definite integral from 1 to sin(x) of (2t - 1) dt.
  • 🤔 The challenge lies in finding F'(x) due to the upper bound being sin(x) instead of a constant.
  • 👉 The fundamental theorem of calculus is not directly applicable due to the variable upper bound.
  • 🌟 Introduce a new function g(x) = sin(x) to help express F(x) in terms of h(x) with a variable upper bound.
  • 📝 F(x) can be written as h(g(x)), which allows us to consider the chain rule for differentiation.
  • 🔄 The chain rule implies that F'(x) = h'(g(x)) * g'(x).
  • 📌 Knowing h'(x) = 2x - 1, we can replace x with g(x) to find F'(x).
  • 🎯 F'(x) will have terms involving 2sin(x) - 1 from the h'(g(x)) part.
  • 🌈 The derivative of g(x) = sin(x) is g'(x) = cos(x), contributing to the F'(x) expression.
  • 🔧 The final expression for F'(x) is 2sin(x) - 1 * cos(x), demonstrating the application of the chain rule.
  • 💡 This process highlights the importance of understanding variable upper bounds and the chain rule in calculus.
Q & A
  • What is the function F(x) as described in the transcript?

    -F(x) is defined as the definite integral of the function 2t - 1 from 1 to sin(x).

  • Why is the upper bound sin(x) significant in the calculation of F'(x)?

    -The upper bound sin(x) is significant because it involves x, thus making F(x) a function of x. This influences how we apply the fundamental theorem of calculus and requires the use of the chain rule to find F'(x).

  • What would F'(x) be if the upper bound were just x instead of sin(x)?

    -If the upper bound were just x, then F'(x) would be straightforwardly calculated as 2x - 1 according to the fundamental theorem of calculus, which states that the derivative of an integral with respect to its upper limit is simply the integrand evaluated at that limit.

  • How is the chain rule applied in the context of the function F(x) as described?

    -The chain rule is applied by defining g(x) = sin(x) and recognizing that F(x) = h(g(x)), where h(x) is the integral from 1 to x of 2t - 1 dt. Thus, F'(x) is h'(g(x)) multiplied by g'(x), where g'(x) is the derivative of sin(x), which is cos(x).

  • What is the derivative of the function h(x) as used in the transcript?

    -The derivative of h(x), where h(x) is the integral from 1 to x of 2t - 1 dt, is 2x - 1.

  • What is g(x) and g'(x) in the context of finding F'(x)?

    -In the transcript, g(x) is defined as sin(x). Therefore, g'(x), the derivative of sin(x), is cos(x).

  • How does substituting g(x) for x affect the calculation of F'(x)?

    -Substituting g(x) for x in h(x) to get h(g(x)) allows the application of the chain rule. The derivative of h with respect to x becomes the derivative of h with respect to g(x), evaluated at sin(x), and then multiplied by the derivative of g, which is cos(x).

  • What is the final expression for F'(x) according to the script?

    -The final expression for F'(x) is (2 sin(x) - 1) multiplied by cos(x).

  • How would one simplify the expression for F'(x)?

    -To simplify the expression for F'(x), one could distribute the cos(x) across the terms inside the parenthesis, resulting in (2 sin(x) cos(x) - cos(x)).

  • What are the practical applications of knowing how to differentiate a function like F(x)?

    -Knowing how to differentiate a function like F(x) is crucial in fields such as physics, engineering, and economics where understanding the rate of change of a function with respect to its variables is needed for modeling and solving real-world problems.

Outlines
00:00
🧮 Understanding the Derivative of a Defined Function

The video begins by introducing a function, F(x), defined as the definite integral from 1 to sine(x) of 2t - 1 dt. The challenge is to find the derivative F'(x). The instructor suggests pausing to think about the solution. The fundamental theorem of calculus is mentioned, which would simplify the process if the upper bound was x instead of sine(x). The concept of g(x) = sine(x) is introduced to apply the chain rule, transforming the problem into finding the derivative of h(g(x)) where h(x) is the integral from 1 to x of 2t - 1 dt. The derivative h'(x) is known, and by applying the chain rule, F'(x) is found to be h'(g(x)) times g'(x), which simplifies to 2 sine(x) - 1 times cosine(x).

Mindmap
Keywords
💡Definite Integral
A definite integral is a fundamental concept in calculus that represents the area under a curve between two points on the graph of a function. In the video, the function F of x is defined as a definite integral from 1 to sine of x of the function two t minus one dt, which is used to introduce the main problem of finding the derivative of F.
💡Upper Bound
In the context of integration, the upper bound refers to the higher limit of the interval over which the integral is calculated. In the video, the upper bound is not a constant x but a function of x, specifically sine of x, which complicates the application of the fundamental theorem of calculus.
💡Fundamental Theorem of Calculus
The fundamental theorem of calculus is a key theorem that links the concept of integration with differentiation. It states that if a function h of x is defined as the definite integral of a function from a to x, then the derivative of h, h prime of x, is equal to the integrand evaluated at x. The video discusses how this theorem would simplify the problem if the upper bound were simply x instead of sine of x.
💡Inner Function
In the context of the fundamental theorem of calculus, the inner function refers to the function that is being integrated. In the video, the inner function is two t minus one, which is the function being integrated from 1 to x in the hypothetical function h of x.
💡Chain Rule
The chain rule is a basic theorem in calculus that is used to compute the derivative of a composite function. 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 find the derivative of F of x by considering F as a composite function h of g of x, where g of x is sine of x.
💡Composite Function
A composite function is a function that is formed by applying one function to the result of another. In the video, the function F of x is expressed as a composite function h of g of x, where g of x is sine of x. This expression is crucial for applying the chain rule to find the derivative of F.
💡Derivative
A derivative in calculus represents the rate at which a function changes with respect to a change in its variable. The video focuses on finding the derivative of the function F of x, denoted as F prime of x, which is the main objective of the problem presented.
💡Sine Function
The sine function is a trigonometric function that describes certain periodic phenomena. In the video, the sine function is used as the upper bound in the integral that defines the function F of x, and later as the argument for the function g of x in the application of the chain rule.
💡Cosine Function
The cosine function is another trigonometric function that is closely related to the sine function. In the video, the derivative of the sine function, which is the cosine function, is used to find the derivative of g of x, which is a part of the chain rule application to find F prime of x.
💡Integration by Substitution
Integration by substitution is a method used to evaluate integrals that involve composite functions. Although not explicitly mentioned in the video, the concept is analogous to the approach taken when defining g of x and then applying the chain rule to find the derivative of F of x.
💡Differentiation
Differentiation is the process of finding the derivative of a function. It is the inverse process of integration and is the main focus of the video as the instructor guides the viewer through finding the derivative of the function F of x.
Highlights

The function F(x) is defined as the definite integral from 1 to sine of x of (2t - 1) dt.

The goal is to find the derivative F'(x) of the function F(x).

The fundamental theorem of calculus is introduced as a potential tool for finding the derivative.

A hypothetical function h(x) is used to illustrate the application of the fundamental theorem of calculus.

The challenge arises due to the upper bound of the integral being sine of x instead of a constant x.

The concept of g(x) is introduced as a substitution to handle the non-constant upper bound.

F(x) is expressed in terms of h(g(x)) to apply the chain rule for differentiation.

The chain rule is applied to find F'(x) as h'(g(x)) * g'(x).

h'(x) is identified as 2x - 1, which will be used in the derivative of F(x).

The derivative of g(x), which is sine of x, is found to be cosine of x.

F'(x) is derived as 2 * sine of x - 1 multiplied by cosine of x.

The process simplifies the derivative by substituting g(x) with sine of x and g'(x) with cosine of x.

The final expression for F'(x) is obtained by combining the results from the chain rule application.

The method demonstrates the use of substitution and the chain rule for derivatives of composite functions.

The video challenges the viewer to pause and attempt the problem before revealing the solution.

The video provides a step-by-step approach to solving the problem, enhancing understanding.

The video concludes with a simplified expression for the derivative, making it accessible for further study.

Transcripts
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