Friday
December 19, 2014

Homework Help: // calculus - derivatives

Posted by Laura on Saturday, October 24, 2009 at 5:56pm.

y = sec(5x) / x^2

my notes say y' = 5x^2sec5xtan5x-2xsec5x
i get y'= sec5x(-2x^-3)+sec5xtan5x(5)(x^-2) when i change it to (sec5x)(x^-2)
or by quotient rule
[ (x^2)(sec5x)(tan5x)(5)-(sec5x)(2x) ] / (x^4)
which is correct?

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