Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Problem Set 2, Integrations techniques, Problem 5A-3g. Can someone please explain this in layman's terms?

OCW Scholar - Single Variable Calculus
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

Problem Set 2, Intgegrations techniques, Problem 5A-3h. I am having the same problem. I can't seem to understand what the solution is trying to explain. Please advise.
The first line is stating known facts about the function inside of the inverse tangent (3g), using y as a shorthand for the function to make the math a little more intuitive. When we rewrite it as \[\tan^{-1} y\]it becomes much more intuitive to write the derivative of the function as \[\ \frac{ \frac{ dy }{ dx } }{ 1+y^2 }\]. We have to have the dy/dx in the top because y is still a function of x, and we are taking the derivative with respect to x, so that function has to have the derivative taken according to the chain rule, and while we could substitute in y in the bottom, it makes it much easier to manipulate our original y to fit that denominator in this case, hence\[y=\frac{ x }{ \sqrt{1-x^2}}\]\[y^2=\frac{ x^2 }{ 1-x^2 }\]\[1+y^2=\frac{ x^2 }{ 1-x^2 }+1\]which we then have to get a similar denominator to add these, which will finally yield the stated equation on the first line of the solution\[1+y^2=\frac{ 1 }{ 1-x^2 }\]From there, we just substitute everything in to the derivative, to get the stated intermediate step, and simplify. Hope that helps for 3g, let me know if not and I will try to write it out better for you. 3h uses a similar technique, I could write it out for you if you need still, but I think this solution and that one are extremely similar.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Not the answer you are looking for?

Search for more explanations.

Ask your own question