Give the Indicated derivative: f'(e) where f(x)=tan^-1(lnx)

Answer 1

f(x) = arctan(lnx)

f'(x)= 1/1+ln^2(x)•1/x = 1/(x)(1+ln^2(x))

f'(e)= 1/(e)(1+ln^2(e)) = 1/(e)(1+1) = 1/2e

Answer 2

f'(x) = 1/(1 + ln²x)·1/x; f'(e) = 1/(1 + 1)·1/e = 1/(2e)

Answer 3

f(x) = tan^-1(lnx) = arctan(lnx)

(derivative of arctanx = 1/(1+x^2, use the chain rule. u=lnx)

f'(x) =[(1/(1+ln^2(x))](lnx)'

(lnx)'=1/x

f'(x) = (1/x)/(ln^2(x)+1)

f'(x)= 1/(x[ln^2(x))+1]

f'(e) = 1/e(ln^2(e) +1)

(lne=1)

=1/(e(1^2+1)

= 1/e(1+1)

f'(x) = 1/2e