What is the implicit derivative of #5=-yx^2-xy/(y-1)+y^2x#?

Question

Evelyn Agard · Accepted Answer

#(dy)/(dx)=(2xy+y/(y-1)-y^2)/(-x^2-x/(y-1)^2+2yx)# First we take the change in #x# of both sides. #d/dx[5]=d/dx[-yx^2]-d/dx[(xy)/(y-1)]+d/dx[y^2x]# #0=-yd/dx[x^2]+x^2d/dx[-y]-y/(y-1)d/dx[x]+xd/dx[y/(y-1)]+y^2d/dx[x]+xd/dx[y^2]# #0=-y(2x)+x^2d/dx[-y]-y/(y-1)(1)+xd/dx[y/(y-1)]+y^2(1)+xd/dx[y^2]# #0=-2xy+x^2d/dx[-y]-y/(y-1)+xd/dx[y/(y-1)]+y^2+xd/dx[y^2]# The chain rule tells us that #d/dx=d/dyxx(dy)/(dx)#. That gives us: #0=-2xy+x^2(dy)/(dx)d/dy[-y]-y/(y-1)+x(dy)/(dx)d/dy[y/(y-1)]+y^2+x(dy)/(dx)d/dy[y^2]# #0=-2xy+(dy)/(dx)(-x^2)-y/(y-1)+(dy)/(dx)x(((y-1)d/dy(y)-(yd/dy(y-1)))/(y-1)^2)+y^2+(dy)/(dx)(2yx)# #0=-2xy+(dy)/(dx)(-x^2)-y/(y-1)+(dy)/(dx)x(((y-1)(1)-y(1))/(y-1)^2)+y^2+(dy)/(dx)(2yx)# #0=-2xy+(dy)/(dx)(-x^2)-y/(y-1)+(dy)/(dx)x((y-1-y)/(y-1)^2)+y^2+(dy)/(dx)(2yx)# #0=-2xy+(dy)/(dx)(-x^2)-y/(y-1)+(dy)/(dx)x((-1)/(y-1)^2)+y^2+(dy)/(dx)(2yx)# #0=-2xy+(dy)/(dx)(-x^2)-y/(y-1)+(dy)/(dx)(-x)/(y-1)^2+y^2+(dy)/(dx)(2yx)# #(dy)/(dx)(-x^2)+(dy)/(dx)(-x)/(y-1)^2+(dy)/(dx)(2yx)=2xy+y/(y-1)-y^2# #(dy)/(dx)(-x^2-x/(y-1)^2+2yx)=2xy+y/(y-1)-y^2# #(dy)/(dx)=(2xy+y/(y-1)-y^2)/(-x^2-x/(y-1)^2+2yx)#

Evelyn Agard · Answer

undefined To find the implicit derivative of $ 5 = -yx^2 - \frac{xy}{y - 1} + y^2x $, differentiate both sides of the equation with respect to $ x $ using the chain rule and product rule where necessary. The implicit derivative is:

$$
\frac{dy}{dx} = \frac{2xy^2 - y^3 - y^2 + 1}{y^2 - 2xy - x(y - 1)^2}
$$