How do you find the derivative of #(cosx)^tanx#?

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Answer 1 · 2017-02-19T20:01:56

#d/dx ((cosx)^(tanx)) = (cosx)^(tanx) (( ln(cosx)-sin^2x)/cos^2x)#

Write the function as:

#(cosx)^(tanx) = (e^(ln(cosx)))^(tanx) = e^(ln(cosx)tanx)#

Now differentiate using the chain rule:

#d/dx (e^(ln(cosx)tanx)) = e^(ln(cosx)tanx) d/dx(ln(cosx)tanx)#

#d/dx (e^(ln(cosx)tanx)) = e^(ln(cosx)tanx) (tanx d/dx(ln(cosx))+ ln(cosx)d/dx(tanx))#

#d/dx (e^(ln(cosx)tanx)) = e^(ln(cosx)tanx) (tanx (-sinx/cosx)+ ln(cosx)/cos^2x)#

#d/dx (e^(ln(cosx)tanx)) = (cosx)^(tanx) ( ln(cosx)-sin^2x)/cos^2x#