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

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

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