# How do you implicitly differentiate #y= y(x-y)^2 + e^(x y) #?

To differentiate the R.H.S., we use the Product Rule and the

Chain Rule :

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To implicitly differentiate ( y = (x - y)^2 + e^{xy} ) with respect to ( x ), you apply the chain rule and product rule where necessary.

[ \begin{align*} y &= (x - y)^2 + e^{xy} \ \frac{dy}{dx} &= \frac{d}{dx}((x - y)^2) + \frac{d}{dx}(e^{xy}) \ &= 2(x - y) \cdot \frac{d}{dx}(x - y) + e^{xy} \cdot \frac{d}{dx}(xy) \ &= 2(x - y) \cdot (1 - \frac{dy}{dx}) + e^{xy} \cdot (x \cdot \frac{dy}{dx} + y) \ &= 2(x - y) - 2(x - y)\frac{dy}{dx} + xe^{xy}\frac{dy}{dx} + e^{xy}y \end{align*} ]

Now, solve for ( \frac{dy}{dx} ):

[ \begin{align*} \frac{dy}{dx} &= \frac{2(x - y) - e^{xy}y}{2(x - y) + xe^{xy}} \ &= \frac{2x - 2y - e^{xy}y}{2x - 2y + xe^{xy}} \end{align*} ]

This is the implicit derivative of ( y ) with respect to ( x ).

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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.

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.

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.

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