How do you do implicit differentiation for #x^2y^2+xy=2#?

Answer 1

Try this (remember that #y# is function of #x#):

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Answer 2

To perform implicit differentiation for (x^2y^2 + xy = 2), differentiate each term with respect to (x) using the chain rule for terms involving (y). Then, solve for (\frac{{dy}}{{dx}}).

Starting with (x^2y^2 + xy = 2): [2x \cdot y^2 + x^2 \cdot 2y \cdot \frac{{dy}}{{dx}} + y^2 + x \cdot \frac{{dy}}{{dx}} = 0] [2xy^2 + 2x^2y\frac{{dy}}{{dx}} + y^2 + xy\frac{{dy}}{{dx}} = 0] [2x^2y\frac{{dy}}{{dx}} + xy\frac{{dy}}{{dx}} = -2xy^2 - y^2] [y\frac{{dy}}{{dx}}(2x^2 + x) = -y^2(2x + 1)] [\frac{{dy}}{{dx}} = \frac{{-y^2(2x + 1)}}{{y(2x^2 + x)}}] [\frac{{dy}}{{dx}} = \frac{{-2xy^2 - y^2}}{{2x^2y + xy}}] [\frac{{dy}}{{dx}} = \frac{{-y(2x + 1)}}{{x(2x + 1)}}]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.