How do you differentiate $y=sec^-1(e^(2x))$ ?

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Answer 1 · 2018-02-18T03:33:17

$2/sqrt{e^{4x}-1}$

$sec y = e^{2x}$

differentiating both sides with respect to $x$ :

$sec y tan y {dy}/{dx}=2e^{2x} implies$

$dy/dx = {2 e^{2x}}/{sec y tan y}={2e^{2x}}/{e^{2x}sqrt{(e^{2x})^2-1}}=2/sqrt{e^{4x}-1}$

How do you differentiate y=sec^-1(e^(2x))y=sec^-1(e^(2x))?