How do you implicitly differentiate # 2x/y = ysqrt(x^2+y^2)-x#?

1 Answer
Apr 24, 2018

#dy/dx=-(yx(x^2+y^2)^(-1/2)-1-2y^-1)/(xy^-2-(x^2+y^2)^(1/2)+y^2(x^2+y^2)^(-1/2))#

Explanation:

Alright, this is a very long one. I'll number each step to make it easier, and also I didn't combine steps so you knew what was going on.

  1. Start with:
    #2xy^-1=y(x^2+y^2)^(1/2)-x#

First we take #d/dx# of each term:
2. #d/dx[2xy^-1]=d/dx[y(x^2+y^2)^(1/2)]-d/dx[x]#
3. #d/dx[2x]y^-1+xd/dx[y^-1]=d/dx[y] (x^2+y^2)^(1/2)+yd/dx[(x^2+y^2)^(1/2)]-d/dx[x]#
4. #2y^-1+xd/dx[y^-1]=d/dx[y] (x^2+y^2)^(1/2)+(y(x^2+y^2)^(-1/2))/2d/dx[x^2+y^2]-1#
5. #2y^-1+xd/dx[y^-1]=d/dx[y] (x^2+y^2)^(1/2)+(y(x^2+y^2)^(-1/2))/2(d/dx[x^2]+d/dx[y^2])-1#
6. #2y^-1+xd/dx[y^-1]=d/dx[y] (x^2+y^2)^(1/2)+(y(x^2+y^2)^(-1/2))/2(2x+d/dx[y^2])-1#

Now we use #d/dx=d/dy*dy/dx#:
7. #2y^-1-dy/dxxy^-2=dy/dx(x^2+y^2)^(1/2)+(y(x^2+y^2)^(-1/2))/2(2x+dy/dx2y)-1#
8. Now we rearrange:
#-dy/dx(xy^-2-(x^2+y^2)^(1/2))=yx(x^2+y^2)^(-1/2)+dy/dxy^2(x^2+y^2)^(-1/2)-1-2y^-1#
9. #-dy/dx(xy^-2-(x^2+y^2)^(1/2)+y^2(x^2+y^2)^(-1/2))=yx(x^2+y^2)^(-1/2)-1-2y^-1#
10. #dy/dx=-(yx(x^2+y^2)^(-1/2)-1-2y^-1)/(xy^-2-(x^2+y^2)^(1/2)+y^2(x^2+y^2)^(-1/2))#