Implicit differentiation

**Jon123** · May 9th 2011, 06:27 AM

Hi, please help!

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!

**Sambit** · May 9th 2011, 06:50 AM

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.

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