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

Question

1291 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-29T16:14:03

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

Firstly multiply everything by #" "(y-x)" "# to remove the award denominator.

assuming #y!=x#

#4(y-x)=y(y-x)-e^(2y)#

#4y-4x=y^2-xy-e^(2y)#

now differentiate.

#d/(dx)(4y-4x)=d/(dx)(y^2-xy-e^(2y))#

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

now rearrange.

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

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

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