Use implicit differentiation to find a formula for dy dx.

$$xy^2-3ye^x=x-3$$

Question

Use implicit differentiation to find a formula for dy dx.

$$xy^2-3ye^x=x-3$$

stamp · Answer

$$xy^2-3ye^x=x-3$$

stamp · Answer

been a while since I've done implicit differentiation so I'm a bit hazy

loser66 · Answer

just derivative both sides

stamp · Answer

$$d(xy^2-3ye^x)=1$$

loser66 · Answer

(y(xy-3e^x))'=1, right? do as usual

loser66 · Answer

hey, I lost hihihihi... please take steps

stamp · Answer

brb I need water/air haha

loser66 · Answer

$$y'(xy-3e^x)+y(xy-3e^x)'=1$$
$$(y'xy -y'3e^x)+y(y+xy'-3e^x)=1$$
$$y'xy -y'3e^x+y^2+yxy'-y3e^x=1$$
factor y'
$$y'(2xy-3e^x)+y^2-y3e^x =1$$
$$y'=\dfrac{1-y^2+3ye^x}{2xy-3e^x}$$

stamp · Answer

Alright I'm gonna try and replicate that without looking on my paper, then look to see if I did the right thing

loser66 · Answer

good guy,

stamp · Answer

Your answer is correct, and I'm close to it all clicking. I'm going to close this question because the rest is on me and I can feel it, but thank you.

stamp · Answer

$$x d(y^2)+y^2d(x)-3yd(e^x)-e^xd(3y)=d(x-3)$$$$y'2xy+y^2-3ye^x-y'3e^x=1$$$$y'(2xy-3e^x)+y^2-3ye^x=1$$$$y'(2xy-3e^x)=1-y^2+3ye^x$$$$y'=\frac{1-y^2+3ye^x}{2xy-3e^x}$$