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

Question

Charles Anctil · Accepted Answer

#(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 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)#

William Abdalla · Answer

undefined To find the derivative of $ (\tan(x))^2 (\sec(x))^5 $, you can use the product rule and the chain rule.

Let $ u = (\tan(x))^2 $ and $ v = (\sec(x))^5 $.

$ \frac{du}{dx} = 2\tan(x) \sec^2(x) $
$ \frac{dv}{dx} = 5(\sec(x))^4 \sec(x)\tan(x) $

Now, apply the product rule:
$ \frac{d}{dx}[(\tan(x))^2 (\sec(x))^5] = u'v + uv' $

Substitute the derivatives:
$ = (2\tan(x) \sec^2(x))(\sec(x))^5 + (\tan(x))^2(5(\sec(x))^4 \sec(x)\tan(x)) $

Simplify:
$ = 2(\tan(x))^2(\sec(x))^5 + 5(\tan(x))^3(\sec(x))^5 $

So, the derivative of $ (\tan(x))^2 (\sec(x))^5 $ is $ 2(\tan(x))^2(\sec(x))^5 + 5(\tan(x))^3(\sec(x))^5 $.