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)
