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bmelyk
Group Title
wondering if someone will walk me through how to do this problem:
 2 years ago
 2 years ago
bmelyk Group Title
wondering if someone will walk me through how to do this problem:
 2 years ago
 2 years ago

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bmelyk Group TitleBest ResponseYou've already chosen the best response.1
Use the appropriate rule or combination of rules to find the derivative of the function defined below. \[y=\sec^2(t^21)\]
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
first can u find the derivative of sec^2 x ?
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.1
well to the power of 2x i guess.
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
to the power of 2x ? u know chain rule, right ? what is derivative of x^2 first ?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
its 2x, right, so derivative of sec^2 x = ?
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.1
2x(sec)*(sec2xtan2x) ??
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
applying chain rule, u get \((sec^2x)' = 2sec\: x(sec\: x)'=2sec \:x(sec\:xtan\:x)\) got this ?
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.1
yes i got it.
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
so now can u apply chain rule again and find it for your question ?
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.1
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)
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
oh, sorry for confusion again....
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.1
haha its okay, theres just no x's present in the question is all.
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.1
so now i have: \[\sec^2(2t)+(t^21)(2)(\sec^2)(sectan)\]
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
how sec^2 2t? the angle doesn't change...
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.1
im using the product rule.
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
\([sec^2(t^21)]'= 2sec (t^21)[sec(t^21)]'\) got this step ?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
now next step would be ?
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.1
to use thechain rule on [sec(t2−1)
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
yes. try it out.
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.1
sectan(t^21)(2t) ??
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
yes! precisely it would be sec(t^21)tan(t^21)(2t)
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
so final answe would be ?
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.1
so now i have: \[2\sec(t^21)*\sec(t^21)\tan(t^21)(2t)\]
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
correct! than can be simplified to \(2\sec^2(t^21)*\tan(t^21)(2t)\)
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
\(4t\sec^2(t^21)*\tan(t^21)\)
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.1
so that's my final answer? : )
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
yup. any doubts ?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
but, did u get all steps ?
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.1
can i write that as: \[4t*\sec(t^21)^2*\tan(t^21)\]
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.1
i wrote them all down : ) i get them now, it's not as hard when i got someone explaining it to me hah.
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.1
can i write it like that though???
 2 years ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.1
should there be a *t at the end of the equation?
 2 years ago
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