Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

whats the derivative to arcsecx? or sec^(-1)x

Mathematics
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

arcsec(x) = 1/(sqrt(1-1/x^2) x^2)
\[\frac{d}{dx} \sec^{-1}(x)=\frac{1}{|x|\sqrt{x^2-1}}\] You can drop the ||'s if x is positive.
let \[y=\sec^{-1}(x)\] so \[\sec(y)=x\] therefore \[y'\sec(y)\tan(y)=1\] => \[y'=\frac{1}{\sec(y)\tan(y)}\] but we need this in terms of x sec(y)=x remember! and we can find tan(y) by looking at what sec(y) means sec(y)=hyp/adj=x/1 so the missing side is opposite to y so we can find it by doing sqrt{x^2-1} so we have \[y'=\frac{1}{x*\frac{\sqrt{x^2-1}}{1}}=\frac{1}{x \sqrt{x^2-1}}\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

by the way tan(y) is opposite/adjacent
very funny
whats funny lol
thanks myininaya
np

Not the answer you are looking for?

Search for more explanations.

Ask your own question