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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?

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1s2s2p Share
Jun 19, 2018

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#

Explanation:

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#

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