How do you implicitly differentiate 4=y-e^(2y)/(y-x)4=ye2yyx?

1 Answer
Nov 29, 2016

(dy)/(dx)=(4-y)/[4-2y+x+2e^(2ydydx=4y42y+x+2e2y

Explanation:

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

assuming y!=xyx

4(y-x)=y(y-x)-e^(2y)4(yx)=y(yx)e2y

4y-4x=y^2-xy-e^(2y)4y4x=y2xye2y

now differentiate.

d/(dx)(4y-4x)=d/(dx)(y^2-xy-e^(2y))ddx(4y4x)=ddx(y2xye2y)

4(dy)/(dx)-4=2y(dy)/(dx)-(y+x(dy)/(dx))-2(dy)/(dx)e^(2y)4dydx4=2ydydx(y+xdydx)2dydxe2y

now rearrange.

4(dy)/(dx)-2y(dy)/(dx)+x(dy)/(dx)+2(dy)/(dx)e^(2y)=4-y4dydx2ydydx+xdydx+2dydxe2y=4y

(dy)/(dx)[4-2y+x+2e^(2y)]=4-ydydx[42y+x+2e2y]=4y

(dy)/(dx)=(4-y)/[4-2y+x+2e^(2ydydx=4y42y+x+2e2y