How do you implicitly differentiate #-y^2=e^(2x-4y)-2yx #?

Question

1449 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-01T12:34:35

#dy/dx=((e^(x-2y))^2-y)/(2(e^(x-2y))^2+x-y)#

We can write this as:
#2yx-y^2=(e^(x-2y))^2#

Now we take #d/dx# of each term:
#d/dx[2yx]-d/dx[y^2]=d/dx[(e^(x-2y))^2]#

#2yd/dx[x]+xd/dx[2y]-d/dx[y^2]=2(e^(x-2y))d/dx[e^(x-2y)]#

#2yd/dx[x]+xd/dx[2y]-d/dx[y^2]=2(e^(x-2y))d/dx[x-2y]e^(x-2y)#

#2yd/dx[x]+xd/dx[2y]-d/dx[y^2]=2(e^(x-2y))e^(x-2y)(d/dx[x]-d/dx[2y])#

#2y+xd/dx[2y]-d/dx[y^2]=2(e^(x-2y))^2(1-d/dx[2y])#

Using the chain rule we get:
#d/dx=dy/dx*d/dy#

#2y+dy/dxxd/dy[2y]-dy/dxd/dy[y^2]=2(e^(x-2y))^2(1-dy/dxd/dy[2y])#

#2y+dy/dx2x-dy/dx2y=2(e^(x-2y))^2(1-dy/dx2)#

#2y+dy/dx2x-dy/dx2y=2(e^(x-2y))^2-dy/dx4(e^(x-2y))^2#

#dy/dx4(e^(x-2y))^2+dy/dx2x-dy/dx2y=2(e^(x-2y))^2-2y#

#dy/dx(4(e^(x-2y))^2+2x-2y)=2(e^(x-2y))^2-2y#

#dy/dx=(2(e^(x-2y))^2-2y)/(4(e^(x-2y))^2+2x-2y)=((e^(x-2y))^2-y)/(2(e^(x-2y))^2+x-y)#