1. ## Implicit differentiation

question is

Find all points on the curve x^2y^2+xy=2 where the slope of the tangent is –1.

d/dx (2x²y) + (2xy²) dy/dx + (y) + (x) dy/dx = 0

Why is " y + x (dy/dx) = 0 " , shouldn't it be " x + y (dy/dx) = 0 " Kindly explain.

dy/dx (2xy² + x) = -2x² y-y
dy/dx = -2x² y-y / 2xy² + x

does it then = x - y / y + x
and does that mean it = -1 because of " = x - y / y + x "

Many thanks for you help if your can explain this!

2. Originally Posted by Jon123
Why is " y + x (dy/dx) = 0 " , shouldn't it be " x + y (dy/dx) = 0 " Kindly explain.
I think you mean the differentiating the part $xy$ w.r.t. $x$.

According to the product rule of differentiation, $\frac{d}{dx}(xy)=x\frac{d}{dx}(y)+y\frac{d}{dx}(x) =x\frac{dy}{dx}+y\frac{dx}{dx}=x\frac{dy}{dx}+y$. (first fix $x$ and differentiate $y$ w.r.t $y$; then fix $y$ and differentiate $x$ w.r.t $y$)

dy/dx (2xy² + x) = -2x² y-y
dy/dx = -2x² y-y / 2xy² + x
You made mistake in this part.

$\frac{d}{dx}(x^2y^2+xy) = 2xy^2+2x^2y\frac{dy}{dx}+x\frac{dy}{dx}+y$.

So $\frac{dy}{dx} = \frac{-y-2xy^2}{x+2x^2y}$. This is your slope. Equate this with $-1$ and simplify to get the answer.