If f(x) = ln (cot e^x), then f '(x) =?

Question

possible book solutions....

a. e^x(cot e^x)

b. -e^x(cot e^x)

c. e^x (sec e^x )(csc e^x)

d. -e^x (sec e^x )(csc e^x)

e. e^x(tan e^x)

f. - e^x(tan e^x)

g. none of these

Answer 1

Use the chain rule twice.

f(x) = ln (cot e^x)

f'(x) = [1/(cot e^x)](-csc² e^x)(e^x)
= -e^x [(sin e^x)/(cos e^x)][1/(sin² e^x)]
= -e^x [1/(cos e^x)] [1/(sin e^x)]
= -e^x (sec e^x) (csc e^x)

The answer is

d. -e^x (sec e^x )(csc e^x).

Answer 2

none of these