# What is the implicit derivative of #1=xsiny#?

Therefore,

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To find the implicit derivative of (1 = x \sin(y)), we can differentiate both sides of the equation with respect to (x) using the chain rule:

[ \frac{d}{dx}(1) = \frac{d}{dx}(x \sin(y)) ]

Since the derivative of a constant is zero, the left side simplifies to (0). For the right side, we apply the product rule:

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

The derivative of (\sin(y)) with respect to (x) involves the chain rule again:

[ \frac{d}{dx}(\sin(y)) = \frac{d\sin(y)}{dy} \cdot \frac{dy}{dx} = \cos(y) \frac{dy}{dx} ]

Substituting this into our expression, we get:

[ x \cos(y) \frac{dy}{dx} + \sin(y) = 0 ]

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

[ \frac{dy}{dx} = -\frac{\sin(y)}{x \cos(y)} ]

Thus, the implicit derivative of (1 = x \sin(y)) with respect to (x) is (-\frac{\sin(y)}{x \cos(y)}).

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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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