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How do you differentiate #f(x)=e^tan(lnx)# using the chain rule?

1 Answer

Oct 29, 2015

#f'(x) = (e^(tan(lnx))sec^2(lnx))/x#

Explanation:

#f'(x) = (e^(tan(lnx)))' = e^(tan(lnx)) * (tan(lnx))' = e^(tan(lnx)) * sec^2(lnx) * (lnx)' = e^(tan(lnx)) * sec^2(lnx) * 1/x#

#f'(x) = (e^(tan(lnx))sec^2(lnx))/x#

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