Respuesta :

meerkat18
y = tan^−1(x2/4) tan(y) = x2/4 sec2(y) = x/2 y′ = xcos^2(y)/2 cos^2(y) = 16x2+16 y′ = 8x/(x2+16)

let u be x2+16
du is 2x dx
dy = 4 du / u
 y = 4 ln (x2 + 16)
 
y at x =0 =  4 ln (16) = 11.09