Find,
$$x^2y^2-2xy+x=2$$
find the slope and the equation of the tangent line at $(2,0)$ by implicit differentiation.
I have gotten this question wrong on a quiz and exam. I'm trying to understand it for my final. Part of the problem is that my text book says find $d/dx$ and then ignore $dy/dx$ after I get
$$x^2y^2=2x^2y$$
Please show me where I am going wrong. I would be able to find the equation of the tangent line if I had the correct $f '(x)$
Not sure what you are doing, but this is my take. Hope it helps. $$ \frac{d}{dx}\left(x^2y^2\right) - \frac{d}{dx}\left(2xy\right) + \frac{d}{dx}\left(x\right) = 0 $$ Then, remember that $\frac{d}{dx}\left(y^2\right)=\frac{d}{dy}\left(y^2\right)\frac{dy}{dx} = 2yy'$ (chain rule). Then, using product rule and chain rule, we have $$ \left(2xy^2 + x^22yy'\right) - \left(2y + 2xy'\right) + 1 = 0 \implies y' = \frac{y-xy^2 + 1/2}{x^2y-x} $$ Look over the Chain rule again, especially when you have $y$ in the expression you are differentiating.