Calculus: Derivative of g(alpha) = 5^(-alpha/2)sin(2*alpha) | Summary and Q&A

399 views
May 16, 2022
by
The Math Sorcerer
YouTube video player
Calculus: Derivative of g(alpha) = 5^(-alpha/2)sin(2*alpha)

TL;DR

Learn how to find the derivative of a function using the product rule and chain rule in calculus.

Install to Summarize YouTube Videos and Get Transcripts

Key Insights

  • 📏 The problem demonstrates the use of the product rule and chain rule in finding the derivative of a complex function.
  • ❓ It highlights the importance of knowing and applying the appropriate formulas in calculus.
  • 💻 The solution showcases the step-by-step process of computing the derivative, considering both the outside and inside functions.
  • 📏 The problem challenges and tests understanding of calculus skills, such as the product rule and the chain rule.

Transcript

in this problem we're going to find the derivative of this function g of alpha equals 5 to the negative alpha over 2 times the sine of 2 alpha solution so we do need a few formulas to do this problem so one formula will be the product rule because we have this function here times this function here there's a quick refresher if you have two function... Read More

Questions & Answers

Q: What formulas are needed to solve this derivative problem?

The product rule and the formula for the derivative of a function with respect to x of a to the x are necessary to solve this problem.

Q: What is the product rule?

The product rule states that when finding the derivative of the product of two functions, we differentiate the first function multiplied by the second function, and then add the first function multiplied by the derivative of the second function.

Q: How does the chain rule apply in this problem?

The chain rule is applied when differentiating the outside function evaluated at the inside function. Here, the outside function is 5 to the negative alpha over 2, and the inside function is negative alpha over 2.

Q: What are the derivatives of the outside and inside functions?

The derivative of the outside function is 5 to the negative alpha over 2 times ln5, and the derivative of the inside function is negative one-half.

Summary & Key Takeaways

  • The problem involves finding the derivative of a function g of alpha equals 5 to the negative alpha over 2 times the sine of 2 alpha.

  • The solution requires using the product rule and the chain rule, along with basic calculus facts.

  • The derivative of the function is computed step-by-step, taking into account the derivatives of the different components.

Share This Summary 📚

Summarize YouTube Videos and Get Video Transcripts with 1-Click

Download browser extensions on:

Explore More Summaries from The Math Sorcerer 📚

Summarize YouTube Videos and Get Video Transcripts with 1-Click

Download browser extensions on: