How do you differentiate #y=sec^-1(4x)#?

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Answer 1 · 2017-04-21T05:51:12

#1/(x sqrt(16x^2 - 1))#

#sec(y(x)) = 4x => d/dx sec(y(x)) = 4#

But

#d/dx sec(y(x)) = sec^2(y(x)) * sin(y(x)) * y'(x)#

So

#y'(x) = 4/(sec^2(y(x)) * sin(y(x)))#

But note that:

#sec(y(x)) = 4x#
#sin(y(x)) = sqrt (1 - cos(y(x))^2) = sqrt (1-(1/(4x))^2)#

Hence

#y'(x) = 4/(sec^2(y(x)) * sin(y(x))) = 1/(x sqrt(16x^2 - 1))#