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

Question

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


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

d.

let p = e^x ; let q = cot e^x = cot p

deriv = d/dx (ln q) = (1/q) dq/dx = (1/cot e^x) d/dx (cot p) =

(1/cot e^x) (-csc^2 p) dp/dx = (1/cot e^x) (-csc^2 e^x) (e^x) =

(sin e^x/cos e^x) (- sin^-2 e^x) (e^x) =

- (e^x) (sin^-1 e^x) (cos^-1 e^x) = -(e^x) (csc e^x) (sec e^x)

Answer 2

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