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March 25, 2017

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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?

  • // calculus - derivatives - ,

    yours is correct

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