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



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bmelyk
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Use the appropriate rule or combination of rules to find the derivative of the function defined below.
\[y=\sec^2(t^21)\]

hartnn
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first can u find the derivative of sec^2 x ?

bmelyk
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sec^2tan^2

bmelyk
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well to the power of 2x i guess.

hartnn
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to the power of 2x ?
u know chain rule, right ?
what is derivative of x^2 first ?

bmelyk
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2x

bmelyk
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1/2rootx

hartnn
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its 2x, right, so derivative of sec^2 x = ?

bmelyk
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2x(sec)*(sec2xtan2x) ??

hartnn
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applying chain rule, u get \((sec^2x)' = 2sec\: x(sec\: x)'=2sec \:x(sec\:xtan\:x)\)
got this ?

bmelyk
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yes i got it.

hartnn
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so now can u apply chain rule again and find it for your question ?

bmelyk
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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
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oh, sorry for confusion again....

bmelyk
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haha its okay, theres just no x's present in the question is all.

bmelyk
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so now i have: \[\sec^2(2t)+(t^21)(2)(\sec^2)(sectan)\]

hartnn
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how sec^2 2t?
the angle doesn't change...

bmelyk
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im using the product rule.

hartnn
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\([sec^2(t^21)]'= 2sec (t^21)[sec(t^21)]'\)
got this step ?

bmelyk
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yup/

hartnn
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now next step would be ?

bmelyk
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to use thechain rule on [sec(t2−1)

hartnn
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yes. try it out.

bmelyk
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sectan(t^21)(2t) ??

hartnn
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yes!
precisely it would be sec(t^21)tan(t^21)(2t)

hartnn
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so final answe would be ?

bmelyk
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so now i have: \[2\sec(t^21)*\sec(t^21)\tan(t^21)(2t)\]

hartnn
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correct!
than can be simplified to \(2\sec^2(t^21)*\tan(t^21)(2t)\)

hartnn
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\(4t\sec^2(t^21)*\tan(t^21)\)

bmelyk
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so that's my final answer? : )

hartnn
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yup.
any doubts ?

hartnn
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but, did u get all steps ?

bmelyk
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can i write that as: \[4t*\sec(t^21)^2*\tan(t^21)\]

bmelyk
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i wrote them all down : ) i get them now, it's not as hard when i got someone explaining it to me hah.

bmelyk
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can i write it like that though???

hartnn
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yup.

bmelyk
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should there be a *t at the end of the equation?