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

1 Answer
May 21, 2016

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

Explanation:

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 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)