How do you differentiate #f(x) =cosx*sec^2(x) #?

Question

Isaiah Allen · Accepted Answer

#d/(d x) f(x)=sec^2 x# #d/(d x) f(x)=-sin x *sec^2 x+ 2secx*sec x*tan x*cos x# #d/(d x) f(x)=-sin x *sec^2 x+ 2sec^2x*tan x*cos x# #d/(d x) f(x)=sec^2(-sin x+tan x*cos x)" , "tan x=(sin x)/(cos x)# #d/(d x) f(x)=sec^2 x(-sin x+(sin x)/cancel(cos x)*cancel(cos x))# #d/(d x) f(x)=sec^2 x(cancel(-sin x)+cancel(sin x))# #d/(d x) f(x)=sec^2 x#

Charles Anctil · Answer

#f'(x) = secxtanx#. Here are two ways to get this answer.

Rewrite first #f(x)=cosxsec^2x# # = cosx 1/(cosx)^2# # = 1/cosx# # = secx#. So, #f'(x) = secxtanx#. Use product rule first #f(x)=cosxsec^2x# #f'(x) = (-sinx)(sec^2x)+(cosx)(2secx*secxtanx)# # = -sinxsec^2x+2sec^2x [tanx cosx]# # = -sinxsec^2x+2sec^2x [sinx]# # = sec^2xsinx# # = secx tanx#

Paisley Johnson · Answer

undefined To differentiate f(x) = cos(x) * sec^2(x), you can use the product rule of differentiation, which states that if you have a function h(x) = u(x) * v(x), then h'(x) = u'(x) * v(x) + u(x) * v'(x). Applying this to f(x), you differentiate each part separately:

1. Differentiate cos(x) with respect to x to get -sin(x).
2. Differentiate sec^2(x) with respect to x to get 2 * sec(x) * tan(x).

Then, apply the product rule:

f'(x) = (-sin(x)) * sec^2(x) + cos(x) * (2 * sec(x) * tan(x))

Simplify further if needed.