# How do you find dy/dx if #x + tan(xy) = 0#?

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To find ( \frac{dy}{dx} ) given ( x + \tan(xy) = 0 ), differentiate both sides of the equation implicitly with respect to ( x ), then solve for ( \frac{dy}{dx} ).

[ \frac{d}{dx}(x) + \frac{d}{dx}(\tan(xy)) = \frac{d}{dx}(0) ]

Using the chain rule for differentiation on ( \tan(xy) ),

[ 1 + \sec^2(xy) \cdot \frac{d(xy)}{dx} = 0 ]

Now, let's apply the product rule for differentiation on ( xy ):

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

Substitute this back into the equation:

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

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

[ \frac{dy}{dx} = -\frac{y\sec^2(xy) + 1}{x\sec^2(xy)} ]

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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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