# How do you implicitly differentiate #2=e^(xy)-xcosy #?

Start from the given equation

differentiate both sides of the equation with respect to x:

solving now for first derivative y'

final answer becomes

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To implicitly differentiate ( 2 = e^{xy} - x \cos(y) ), we follow these steps:

- Differentiate both sides of the equation with respect to ( x ).
- Apply the chain rule and product rule where necessary.
- Solve for ( \frac{{dy}}{{dx}} ).

Starting with the given equation:

[ 2 = e^{xy} - x \cos(y) ]

Differentiating both sides with respect to ( x ):

[ \frac{{d}}{{dx}} (2) = \frac{{d}}{{dx}} (e^{xy} - x \cos(y)) ]

[ 0 = \frac{{d}}{{dx}} (e^{xy}) - \frac{{d}}{{dx}} (x \cos(y)) ]

Using the chain rule for the first term and product rule for the second term:

[ 0 = e^{xy} \left( \frac{{d}}{{dx}} (xy) \right) - \left( \frac{{d}}{{dx}} (x) \cdot \cos(y) + x \cdot \frac{{d}}{{dx}} (\cos(y)) \right) ]

[ 0 = e^{xy} \left( y + x \frac{{dy}}{{dx}} \right) - (1 \cdot \cos(y) + x \cdot (-\sin(y) \cdot \frac{{dy}}{{dx}})) ]

Now, isolate ( \frac{{dy}}{{dx}} ):

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

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

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

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

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To implicitly differentiate (2 = e^{xy} - x \cos y), differentiate each term with respect to (x) using the chain rule for (e^{xy}) and the product rule for (xy).

[\frac{d}{dx}(2) = \frac{d}{dx}(e^{xy}) - \frac{d}{dx}(x\cos y)]

[0 = e^{xy} \frac{d}{dx}(xy) - (\cos y + x\frac{d}{dx}(\cos y))]

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

[0 = ye^{xy} + x^2\frac{dy}{dx}e^{xy} - \cos y + x\sin y\frac{dy}{dx}]

Solve for (\frac{dy}{dx}).

[\frac{dy}{dx}(x^2e^{xy} + x\sin y) = \cos y - ye^{xy}]

[\frac{dy}{dx} = \frac{\cos y - ye^{xy}}{x^2e^{xy} + x\sin 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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