take the second derivative of x^2y^2-2x=3
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Originally Posted by thatonegurl take the second derivative of x^2y^2-2x=3 show us the first derivative ... you need that to get the second.
I'm confused with derivatives all together. As far as the first derivative, I got as far as 2x2yy'-2=0 I'm not sure what to do after that, and I don't even know if that is right.
Originally Posted by thatonegurl I'm confused with derivatives all together. As far as the first derivative, I got as far as 2x2yy'-2=0 I'm not sure what to do after that, and I don't even know if that is right. you need to use the product rule for the first term ... $\displaystyle \frac{d}{dx}(x^2y^2-2x = 3)$ $\displaystyle x^2 \cdot 2y \cdot y' + y^2 \cdot 2x - 2 = 0$ solve for $\displaystyle y'$ , then take the derivative again.
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