F'(x) = 12(x)(arctan (e ^ (2x))) @ ( (1/4)(ln3) , pi (ln3) )

Determine the slope of the tangent line to the graph of

f(x) = 12x * arctan (e ^ (2x)) at ( 1\4 ln (3 , pi ln (3) )

without using a calculator I get

f'(1/4 ln3) = 12arctan ( e^ (ln3 / 2)) + (12ln3e^(ln3/2))/(2+2e(ln3))

is that correct?

let y = 12x * arctan (e ^ (2x))

y ' = [12x /(1 + e^(4x)]*2e^(2x)) + 12 arctan(e^(2x)

=6x e^(-2x) /( 1 + e^(4x) + 12 arctan(e^(2x)

plug in x = (1/4) ln(3)

y ' = 6(1/4)ln 3/√3(1 + 3) + 12 (Π/3)

slope = (√3/8)ln(3) + 4Π