How do you find the derivative of #(tanx)^2 (secx)^5#?

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Answer 1 · 2016-05-21T11:23:43

#(d y)/(d x)=(5sin x tan^2 x)/(cos^6 x)+(2 tan x(1+tan^2 x))/(cos^5 x)#

#y=(tan x)^2*(sec x)^5" "sec x=1/cos x" ; "sec^5 x=1/cos^5 x#

#y=(tan^2 x)/(cos^5 x) #

#(d y)/(d x)=((tan^2 x)^'*cos^5 x-(cos^5 x)^' *tan^2 x)/(cos^5 x)^2#

#(tan^2 x)^'=2 tan x(1+tan ^2 x)#

#(cos^5 x)'=-5sin x cos^4 x#

#(d y)/(d x)=((2 tan x(1+tan ^2 x)*cancel(cos^5 x))-((-5 sin x cancel(cos ^4 x))*tan^2 x))/(cancel(cos^10 x)#

#(d y)/(d x)=((2 tan x(1+tan^2 x)cos x)-((-5sin x)*tan^2 x))/(cos^6 x)#

#(d y)/(d x)=(2tan x(1+tan^2 x)cos x+5sin x tan^2 x)/(cos^6 x)#

#(d y)/(d x)=(5sin x tan^2 x)/(cos^6 x)+(2 tan x(1+tan^2 x)cos x)/(cos^6 x)#

#(d y)/(d x)=(5sin x tan^2 x)/(cos^6 x)+(2 tan x(1+tan^2 x))/(cos^5 x)#