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.

wondering if someone will walk me through how to do this problem:

Calculus1
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

SIGN UP FOR FREE
Use the appropriate rule or combination of rules to find the derivative of the function defined below. \[y=\sec^2(t^2-1)\]
first can u find the derivative of sec^2 x ?
sec^2tan^2

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

well to the power of 2x i guess.
to the power of 2x ? u know chain rule, right ? what is derivative of x^2 first ?
2x
1/2rootx
its 2x, right, so derivative of sec^2 x = ?
2x(sec)*(sec2xtan2x) ??
applying chain rule, u get \((sec^2x)' = 2sec\: x(sec\: x)'=2sec \:x(sec\:xtan\:x)\) got this ?
yes i got it.
so now can u apply chain rule again and find it for your question ?
i was just confused how you just had sec^2x wrote down is all, i thought you meant to the power of 2x, not sec^2(x)
oh, sorry for confusion again....
haha its okay, theres just no x's present in the question is all.
so now i have: \[\sec^2(2t)+(t^2-1)(2)(\sec^2)(sectan)\]
how sec^2 2t? the angle doesn't change...
im using the product rule.
\([sec^2(t^2-1)]'= 2sec (t^2-1)[sec(t^2-1)]'\) got this step ?
yup/
now next step would be ?
to use thechain rule on [sec(t2−1)
yes. try it out.
sectan(t^2-1)(2t) ??
yes! precisely it would be sec(t^2-1)tan(t^2-1)(2t)
so final answe would be ?
so now i have: \[2\sec(t^2-1)*\sec(t^2-1)\tan(t^2-1)(2t)\]
correct! than can be simplified to \(2\sec^2(t^2-1)*\tan(t^2-1)(2t)\)
\(4t\sec^2(t^2-1)*\tan(t^2-1)\)
so that's my final answer? : )
yup. any doubts ?
but, did u get all steps ?
can i write that as: \[4t*\sec(t^2-1)^2*\tan(t^2-1)\]
i wrote them all down : ) i get them now, it's not as hard when i got someone explaining it to me hah.
can i write it like that though???
yup.
should there be a *t at the end of the equation?

Not the answer you are looking for?

Search for more explanations.

Ask your own question