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

1 Answer

Ananda Dasgupta

Feb 18, 2018

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

Explanation:

#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}#

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