anonymous
  • anonymous
wondering if someone will walk me through how to do this problem:
Calculus1
schrodinger
  • schrodinger
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anonymous
  • anonymous
Use the appropriate rule or combination of rules to find the derivative of the function defined below. \[y=\sec^2(t^2-1)\]
hartnn
  • hartnn
first can u find the derivative of sec^2 x ?
anonymous
  • anonymous
sec^2tan^2

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

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