How do you find the derivative of #x^3*arctan(7x)#?

1 Answer
Apr 23, 2015

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)#