Derivative of inverse sine | Taking derivatives | Differential Calculus | Khan Academy

Khan Academy
30 Apr 201404:56
EducationalLearning
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TLDRThe video script explores the process of finding the derivative of the inverse sine function with respect to X. It encourages viewers to pause and attempt the problem independently before revealing the solution. The key steps involve recognizing that the sine of Y equals X, applying implicit differentiation, and using the chain rule. The final result is derived by expressing the derivative in terms of X, leading to the conclusion that the derivative of the inverse sine of X is equal to one over the square root of one minus X squared.

Takeaways
  • 📚 The topic of the video is calculating the derivative of the inverse sine function with respect to X, denoted as Y = sin^(-1)(X).
  • 💡 The video encourages viewers to pause and attempt to solve the problem independently before revealing the solution.
  • 🧩 The first hint suggests using the known derivative of the sine function and applying implicit differentiation to find the derivative of the inverse sine.
  • 🔄 The process starts by setting up the equation sin(Y) = X, which allows for the application of implicit differentiation on both sides with respect to X.
  • 📈 The derivative of sin(Y) with respect to Y is cosine(Y), leading to the equation cos(Y) * dy/dx = d(X)/dx, which simplifies to cos(Y) * dy/dx = 1.
  • 🔧 The video demonstrates the use of the chain rule to find the derivative dy/dx, which is found to be 1/cos(Y) by dividing both sides of the equation by cos(Y).
  • 🔄 The goal is to express the derivative in terms of X, not Y, so the video uses the trigonometric identity sin^2(Y) + cos^2(Y) = 1 to solve for cos(Y) in terms of sin(Y).
  • 🌟 The final expression for the derivative is derived as dy/dx = 1/√(1 - X^2), which is the main result of the video.
  • 📝 The video emphasizes the importance of internalizing mathematical concepts through practice and reproof, suggesting that this knowledge is particularly useful in advanced calculus.
  • 🎓 The video serves as a helpful resource for those learning calculus, especially in understanding and applying the derivative of the inverse sine function.
Q & A

  • What is the main topic of the video?

    -The main topic of the video is to find the derivative of Y with respect to X when Y is the inverse sine of X.

  • What is the first hint given in the video?

    -The first hint is that we don't know the derivative of the inverse sine of X, but we do know the derivative of the sine function.

  • What is the method suggested in the video to find the derivative?

    -The method suggested is to use implicit differentiation to find the derivative of Y with respect to X.

  • What is the relationship established between Y and X in the video?

    -The relationship established is that if Y is the inverse sine of X, then sine of Y is equal to X.

  • How is the derivative of the left-hand side of the equation determined?

    -The derivative of the left-hand side is determined by applying the chain rule, which results in the cosine of Y times the derivative of Y with respect to X.

  • What is the derivative of X with respect to X?

    -The derivative of X with respect to X is 1, as it is a constant with respect to X.

  • How is the final expression for dy/dx derived?

    -The final expression for dy/dx is derived by dividing both sides of the equation by the cosine of Y and using the trigonometric identity to express the cosine of Y in terms of X.

  • What is the final expression for the derivative of the inverse sine of X?

    -The final expression for the derivative of the inverse sine of X is 1 over the square root of 1 minus X squared.

  • Why is it useful to know the derivative of the inverse sine function?

    -It is useful to know the derivative of the inverse sine function as it may appear in more advanced calculus problems and expressions, providing a quick reference for solving related problems.

  • How can one remember the derivative of the inverse sine function?

    -One can remember the derivative of the inverse sine function by practicing its derivation and understanding the steps involved, as well as recognizing its pattern in trigonometric functions.

Outlines
00:00
📚 Derivative of Inverse Sine (arcsin) with Respect to X

This paragraph delves into the process of determining the derivative of the inverse sine function with respect to X. It begins by posing the question and encouraging viewers to attempt the problem independently before providing two hints. The first hint suggests using the known derivative of the sine function and applying implicit differentiation, while the second hint involves rearranging the equation using the chain rule. The explanation continues with the application of these hints, leading to the conclusion that the derivative of Y with respect to X, where Y is the inverse sine of X, is equal to 1 over the cosine of Y. The paragraph then further refines this result by expressing it in terms of X, using trigonometric identities to replace cosine of Y with one minus sine squared of Y. The final result is that the derivative with respect to X of the inverse sine of X is equal to one over the square root of one minus X squared. The paragraph emphasizes the importance of understanding and being able to derive this result as it may prove useful in more advanced calculus studies.

Mindmap
Keywords
💡Derivative
A derivative in calculus represents the rate at which a function is changing at any given point, essentially measuring how a function's output changes as its input changes. In the video, the focus is on finding the derivative of the inverse sine of x with respect to x, demonstrating a fundamental operation in calculus. This concept is central as it transitions from a general curiosity about the derivative of an arcane function to a specific and tangible outcome.
💡Inverse Sine
Inverse sine, denoted as sin^-1(x) or arcsin(x), is the function that reverses the action of the sine function, answering the question 'what angle has sine equal to x?'. In the context of the video, y is defined as the inverse sine of x, setting up a scenario to explore how changes in x affect y through differentiation, thus linking trigonometric functions directly to calculus.
💡Implicit Differentiation
Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly solved for one variable in terms of another. The video uses this approach to handle the equation sine of y equals x, since y is not isolated on one side of the equation. This method allows the differentiation of both sides of the equation with respect to x, which is key to finding the derivative of the inverse sine function.
💡Chain Rule
The chain rule is a formula to compute the derivative of the composition of two or more functions. In the video, it's used to differentiate the sine of y with respect to x, by first taking the derivative of sine of y with respect to y, and then multiplying by the derivative of y with respect to x. This illustrates how composite functions are differentiated and is essential for the problem at hand.
💡Cosine
Cosine is a trigonometric function, showing the ratio between the adjacent side and the hypotenuse of a right-angled triangle. In the video, cosine of y emerges during differentiation as the derivative of sine of y with respect to y. The narrative then transitions to solving the original problem by expressing this cosine term in a form that can be related back to x, highlighting the interconnectedness of trigonometric identities in calculus.
💡Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that are true for all values of the involved variables. The video relies on these identities, specifically the Pythagorean identity, to transform the expression involving cosine of y into one involving sine of y (and thus x), demonstrating the practical utility of these identities in solving calculus problems.
💡Square Root
The square root of a number is a value that, when multiplied by itself, gives the original number. In the context of the video, the square root is used to solve for cosine of y in terms of sine of y, facilitating the final expression of the derivative in terms of x. This step is pivotal in translating the derivative from a function of y to a function of x.
💡Pythagorean Identity
The Pythagorean identity is a fundamental trigonometric identity stating that sine squared of an angle plus cosine squared of the same angle equals one. The video uses this identity to express cosine of y in terms of sine of y, which is crucial for re-expressing the derivative in terms of x, illustrating how trigonometric identities can simplify calculus problems.
💡Principal Root
The principal root of a number refers to its non-negative square root. This concept is mentioned in the video when solving for cosine of y, using the principal root to denote the positive square root of 1 minus sine squared of y. This clarification is vital for maintaining the mathematical integrity of the solution process.
💡Expression Simplification
Expression simplification refers to the process of making an expression easier to understand or solve, often by combining like terms or using identities to replace parts of the expression. In the video, this process is exemplified when the derivative of the inverse sine of x is simplified to 1 over the square root of 1 minus x squared, using trigonometric identities and substitution to achieve a more straightforward form. This illustrates the goal of calculus to express complex rates of change in more accessible terms.
Highlights

Exploring the derivative of the inverse sine function with respect to X.

Encouragement for viewers to pause the video and attempt the problem independently.

The first hint suggests using the known derivative of the sine function and implicit differentiation.

The equivalence of Y as the inverse sine of X to sine of Y equals X.

Performing implicit differentiation by taking the derivative of both sides with respect to X.

Applying the chain rule to find the derivative of the left-hand side.

Derivative of the right-hand side being equal to one since the derivative of X with respect to X is one.

Solving for dy/dx by dividing both sides by the cosine of Y.

Re-expressing the derivative in terms of X by using the trigonometric identity sine squared Y + cosine squared Y equals one.

Derivative of Y with respect to X is found to be one over the cosine of Y.

Substituting X back in for sine of Y to express the derivative in terms of X.

The final expression for the derivative is one over the square root of one minus X squared.

The derivative of the inverse sine of X is equal to one over the square root of one minus X squared.

The importance of internalizing mathematical concepts through practice and reproof.

The practical applications of understanding the derivative of the inverse sine function in advanced calculus.

The method presented serves as a foundation for further exploration in calculus and its applications.

Transcripts

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Derivative CalculationInverse Sine FunctionTrigonometric IdentitiesImplicit DifferentiationMathematical TutorialCalculus ConceptsEducational ContentProblem SolvingMathematics EducationOnline Learning