# What is the implicit derivative of #1= ln(xy) #?

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To find the implicit derivative of (1 = \ln(xy)), we differentiate both sides of the equation with respect to (x), using the chain rule and product rule where necessary.

[ \frac{d}{dx}(1) = \frac{d}{dx}(\ln(xy)) ]

Differentiating (1) with respect to (x) gives (0).

For the right side of the equation, using the chain rule, the derivative of (\ln(xy)) with respect to (x) is:

[ \frac{d}{dx}(\ln(xy)) = \frac{1}{xy} \cdot \frac{d(xy)}{dx} ]

Using the product rule for (xy), we have:

[ \frac{d(xy)}{dx} = x\frac{d(y)}{dx} + y\frac{d(x)}{dx} ]

As (\frac{d(x)}{dx} = 1) and (\frac{d(y)}{dx}) represents (\frac{dy}{dx}), substituting these values, we get:

[ \frac{d(xy)}{dx} = x\frac{dy}{dx} + y ]

Therefore, the implicit derivative is:

[ 0 = \frac{1}{xy}(x\frac{dy}{dx} + y) ]

Rearranging this, we get:

[ x\frac{dy}{dx} = -y ]

Finally, dividing both sides by (x) yields the implicit derivative:

[ \frac{dy}{dx} = -\frac{y}{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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