We know that,
#color(red)((1) d/(dX)(lnX)=1/X#
#color(red)((2) d/(dX)(tan^-1X)=1/(1+X^2#
Here,
#y=ln(tan^-1(ex))#
Take, #y=lnuand u=tan^-1(ex)#
#:.(dy)/(du)=1/u and(du)/(dx)=1/(1+(ex)^2)d/(dx)(ex)#
#where,u=tan^-1(ex) and color(orange)(d/(dx)(ex)=e#
#i.e. (dy)/(du)=1/(tan^-1(ex))and(du)/(dx)=e/(1+e^2x^2)#
#"Using "color(blue)"Chain Rule:"#
#color(blue)((dy)/(dx)=(dy)/(du)xx(du)/(dx)#
#:.(dy)/(dx)=1/(tan^-1(ex))xxe/(1+e^2x^2)#
#=>(dy)/(dx)=e/((1+e^2x^2)tan^-1(ex))#
Note:
If #y=ln(tan^-1(e^x)). then ,take,v=e^x=>(dv)/(dx)=color(orange)(d/(dx)(e^x)=e^x#.
So,the answer will be
#=>(dy)/(dx)=e^x/((1+(e^x)^2)tan^-1(e^x))#
