What is the derivative of #tan^-1(x^2 y^5)#?

1 Answer
Apr 23, 2015

Say that #y=arctan(x^2y^5)# and let's find #(dy)/(dx)#. We're going to use the implicit differentiation formula below to solve this problem...

#f(x)*g'(y)(dy)/(dx)+f'(x)*g(y)#

So, let's begin solving the problem...

#tany=x^2y^5#

#sec^2y*(dy)/(dx)=x^2*5y^4*(dy)/(dx)+2x*y^5#

#(tan^2y+1)*(dy)/(dx)=x^2*5y^4*(dy)/(dx)+2x*y^5#

#(x^4y^10+1)*(dy)/(dx)=x^2*5y^4*(dy)/(dx)+2x*y^5#

#(x^4y^10+1)*(dy)/(dx)-(x^2*5y^4)*(dy)/(dx)=2x*y^5#

#(dy)/(dx)*{(x^4y^10+1)-x^2*5y^4}=2x*y^5#

#(dy)/(dx)*(x^4y^10-x^2*5y^4+1)=2x*y^5#

#(dy)/(dx)*(x^2y^4(x^2y^6-5)+1)=2x*y^5#

#(dy)/(dx)=(2x*y^5)/(x^2y^4(x^2y^6-5)+1)#

To spice things up a little we can find the derivative of this function using partial derivatives...

#z=arctan(x^2y^5)#

#tanz=x^2y^5# (we hold the function of y as a constant from here)

#sec^2z*(delz)/(delx)=2xy^5#

#(tan^2z+1)*(delz)/(delx)=2xy^5#

#(x^4y^10+1)*(delz)/(delx)=2xy^5#

#(delz)/(delx)=(2xy^5)/(x^4y^10+1)#

Alternatively...

#z=arctan(x^2y^5)#

#tanz=x^2y^5#

#sec^2z*(delz)/(dely)=5x^2y^4# (we hold the function of x as a constant from here)

#(tan^2z+1)*(delz)/(dely)=5x^2y^4#

#(x^4y^10+1)*(delz)/(dely)=5x^2y^4#

#(delz)/(dely)=(5x^2y^4)/(x^4y^10+1)#