Firstly let's differentiate this function using implicit and logarithmic differentiation:
#q=ln(arctan(7x))#
#e^q=arctan(7x)#
#tan(e^q)=7x#
#e^qsec^2(e^q)*(dq)/(dx)=7#
#arctan(7x)*(tan^2(e^q)+1)(dq)/(dx)=7#
#arctan(7x)*(49x^2+1)(dq)/(dx)=7#
#(dq)/(dx)=7/(arctan(7x)*(49x^2+1)#
Alright, knowing this we can now differentiate #x^3*arctan(7x)# using implicit differentiation and the result above...
#y=x^3*arctan(7x)#
#lny=ln(x^3*arctan(7x))#
#lny=ln(x^3)+ln(arctan(7x))#
#lny=3lnx+ln(arctan(7x))#
#1/y*(dy)/(dx)=3/x+7/(arctan(7x)*(49x^2+1)#
#(dy)/(dx)=y{3/x+(7)/(arctan(7x)*(49x^2+1)}}#
#(dy)/(dx)=x^3*arctan(7x){3/x+7/(arctan(7x)*(49x^2+1)}}#
#(dy)/(dx)=(3x^3*arctan(7x))/x+(7x^3*arctan(7x))/(arctan(7x)*(49x^2+1))#
#(dy)/(dx)=3x^2*arctan(7x)+(7x^3)/(49x^2+1)#
I can also give you an alternative way of finding this derivative, using the product rule...
#y=x^3*arctan(7x)=u*v#
#u=x^3#, therefore #(du)/(dx)=3x^2#
#v=arctan(7x)#
#tanv=7x#
#sec^2v*(dv)/(dx)=7#
#(tan^2v+1)*(dv)/(dx)=7#
#(49x^2+1)*(dv)/(dx)=7#
#(dv)/(dx)=7/(49x^2+1)#
This means that:
#(dy)/(dx)=x^3*7/(49x^2+1)+arctan(7x)*3x^2#
#(dy)/(dx)=(7x^3)/(49x^2+1)+3x^2*arctan(7x)#
