Having trouble with this problem. t(x) = ln (sec e^2x ). Then t'(x) =?

Question

If this helps, my book gives the following solutions....

a. 2e^2x cot e^2x
b. -2e^2x tan e^2x
c. 2e^2x (sec e^2x )(csc e^2x )
d. -2e^2x (sec e^2x )(csc e^2x )
e. -2e^2x cot e^2x
f. 2e^2x tan e^2x
g. none of these

Answer 1

2e^2xtane^2x

Answer 2

t(x) = ln (sec (e^(2x)))

t'(x) = { 1 / (sec (e^(2x)))] } { sec (e^(2x)) tan (e^(2x)) } {e^(2x)} {2}

= cos(e^(2x)) sec (e^(2x)) tan (e^(2x)) e^(2x) (2)

The cosine and the cosecant will cancel each other out, being reciprocals and all.

= 2e^(2x)tan(e^(2x))

Answer 3

think it is F since derivative of ln(u)=u'/u

derivative of (sec e^2x)/(sec e^2x)