How do you find #(delf(x,y))/(dely)# and #(delf(x,y))/(delx)# of #f(x,y)=(yx^2-2y)/(2ye^x+y^-3)#, using the quotient rule?

Question

Charles Anctil · Accepted Answer

#(delf(x,y))/(dely)=((2ye^x+y^-3)(x^2-2)-(yx^2-2y)(2e^x-3y^-4))/(2ye^x+y^-3)^2#

#(delf(x,y))/(delx)=(2yx(2ye^x+y^-3)-2ye^x(yx^2-2y))/(2ye^x+y^-3)^2# To partially differentiate, we treat one variable as a constant. So, for #(delf(x,y))/(dely)# we treat #x# as a constant. #del/(delx)u/v=(v(delu)/(delx)-u(delv)/(delx))/v^2# #del/(dely)u/v=(v(delu)/(dely)-u(delv)/(dely))/v^2# #u=yx^2-2y# #(delu)/(dely)=x^2-2# #v=2ye^x+y^-3# #(delv)/(dely)=2e^x-3y^-4# #(delf(x,y))/(dely)=((2ye^x+y^-3)(x^2-2)-(yx^2-2y)(2e^x-3y^-4))/(2ye^x+y^-3)^2# Now, for #(delf(x,y))/(delx)# we treat #y# as a constant. #u=yx^2-2y# #(delu)/(delx)=2yx# #v=2ye^x+y^-3# #(delv)/(delx)=2ye^x# #(delf(x,y))/(dely)=(2yx(2ye^x+y^-3)-2ye^x(yx^2-2y))/(2ye^x+y^-3)^2#

Ezra Adams · Answer

undefined To find $ \frac{\partial f(x,y)}{\partial y} $ and $ \frac{\partial f(x,y)}{\partial x} $ of $ f(x,y) = \frac{yx^2 - 2y}{2ye^x + y^{-3}} $ using the quotient rule, follow these steps:

1. For $ \frac{\partial f(x,y)}{\partial y} $:
   Apply the quotient rule, which states that for functions $ u(x) $ and $ v(x) $, the derivative of $ \frac{u(x)}{v(x)} $ is given by $ \frac{u'v - uv'}{v^2} $.
   Let $ u(y) = yx^2 - 2y $ and $ v(y) = 2ye^x + y^{-3} $.
   Find $ \frac{\partial u}{\partial y} $ and $ \frac{\partial v}{\partial y} $.
   Then apply the quotient rule formula.

2. For $ \frac{\partial f(x,y)}{\partial x} $:
   Follow similar steps but this time differentiate with respect to $ x $.

Here are the steps in detail:

1. For $ \frac{\partial f(x,y)}{\partial y} $:
   $ u(y) = yx^2 - 2y $
   $ v(y) = 2ye^x + y^{-3} $
   $ \frac{\partial u}{\partial y} = x^2 - 2 $
   $ \frac{\partial v}{\partial y} = 2e^x - (-3)y^{-4} $
   Apply the quotient rule:
   $ \frac{\partial f}{\partial y} = \frac{(x^2 - 2)(2ye^x + y^{-3}) - (yx^2 - 2y)(2e^x - (-3)y^{-4})}{(2ye^x + y^{-3})^2} $

2. For $ \frac{\partial f(x,y)}{\partial x} $:
   $ u(y) = yx^2 - 2y $
   $ v(y) = 2ye^x + y^{-3} $
   $ \frac{\partial u}{\partial x} = 2yx $
   $ \frac{\partial v}{\partial x} = 2ye^x $
   Apply the quotient rule:
   $ \frac{\partial f}{\partial x} = \frac{(2yx)(2ye^x + y^{-3}) - (yx^2 - 2y)(2ye^x)}{(2ye^x + y^{-3})^2} $