How do you differentiate #1=e^(xy)/(e^x+e^y)#?
i think you should help yourself by simplifying first
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To differentiate the expression (1 = \frac{e^{xy}}{e^x + e^y}), you can use implicit differentiation. The steps are as follows:
 Take the natural logarithm (ln) of both sides.
 Differentiate implicitly with respect to (x).
 Solve for (\frac{dy}{dx}) (the derivative of (y) with respect to (x)).
Here are the detailed steps:

Take ln of both sides: [\ln(1) = \ln\left(\frac{e^{xy}}{e^x + e^y}\right)] [\Rightarrow 0 = \ln(e^{xy})  \ln(e^x + e^y)]

Apply the logarithm properties: [0 = xy  \ln(e^x + e^y)]

Differentiate both sides with respect to (x): [0 = x \frac{dy}{dx} + y  \frac{e^x}{e^x + e^y} \cdot (1 + \frac{dy}{dx})]

Solve for (\frac{dy}{dx}): [\frac{dy}{dx}(x  \frac{e^x}{e^x + e^y}) = y + \frac{e^x}{e^x + e^y}] [\frac{dy}{dx} = \frac{y + \frac{e^x}{e^x + e^y}}{x  \frac{e^x}{e^x + e^y}}]
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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