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)