Implicitly differentiate 3y + 1 = (y - 1) x + y^2 (x - 2)

Question

Implicitly differentiate  3y + 1 = (y - 1) x + y^2 (x - 2)

anonymous · Answer

$$3y + 1 = (y - 1) x + y^2 (x - 2)$$

anonymous · Answer

Distribute the RHS to get:$$3y + 1 = yx - x + y^2x - 2y^2$$Differentiate both sides with respect to x. Remember the product rule, we are left with:$$3y \prime + 1 = y - y \prime x - 1 + y^2 + 2yy \prime x - 4yy \prime$$

anonymous · Answer

Oops, without the +1 on the LHS.

anonymous · Answer

Separate all terms with y' on the LHS and we get:$$y \prime (3 + 4y + x  - 2yx) = y - 1 + y^2$$Divide both sides by (3 + 4y + x - 2yx) and we arrive at:$$y \prime = \frac{y^2 + y - 1}{-2yx + x + 4y + 3}$$

anonymous · Answer

Small sign error, it should be:$$ y \prime = - \frac{y^2+y−1}{2xy+x−4y−3}$$

anonymous · Answer

ok got close justa few sign errors myself... but is that all I feel like I am missing something at the end

anonymous · Answer

Nope, that's the answer, unless you want to factor out y on top or try to do some algebra. The y' is a function of (y,x) when defined implicitly.

anonymous · Answer

ok well thanks you were a huge help!!!

anonymous · Answer

No problem :-)