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Use the appropriate rule or combination of rules to find the derivative of the function defined below. \[y=\sec^2(t^2-1)\]
first can u find the derivative of sec^2 x ?
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 ?
its 2x, right, so derivative of sec^2 x = ?
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 ?
now next step would be ?
to use thechain rule on [sec(t2−1)
yes. try it out.
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 \(2\sec^2(t^2-1)*\tan(t^2-1)(2t)\)
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???
should there be a *t at the end of the equation?