To differentiate $e^{\cos y} = xy$ implicitly with respect to $x$, we can proceed as follows:

1. Differentiate both sides with respect to $x$:

   $\frac{d}{dx}\left(e^{\cos y}\right) = \frac{d}{dx}(xy)$

2. Differentiate the left side:
   Using the chain rule on $e^{\cos y}$:

   $\frac{d}{dx}\left(e^{\cos y}\right) = e^{\cos y} \cdot \frac{d}{dx}(\cos y) = e^{\cos y} \cdot (-\sin y) \cdot \frac{dy}{dx}$

   So the left side becomes:

   $-e^{\cos y} \sin y \cdot \frac{dy}{dx}$

3. Differentiate the right side:
   Applying the product rule to $xy$:

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

4. Combine results:
   Now we have the equation:

   $-e^{\cos y} \sin y \cdot \frac{dy}{dx} = y + x \frac{dy}{dx}$

5. Solve for $\frac{dy}{dx}$:
   Collect all terms involving $\frac{dy}{dx}$ on one side:

   $-e^{\cos y} \sin y \cdot \frac{dy}{dx} - x \frac{dy}{dx} = y$

   Factor out $\frac{dy}{dx}$:

   $\frac{dy}{dx} \left(-e^{\cos y} \sin y - x\right) = y$

   Then, solve for $\frac{dy}{dx}$:

   $\frac{dy}{dx} = \frac{y}{-e^{\cos y} \sin y - x}$

So, the implicit derivative is:

$\frac{dy}{dx} = \frac{y}{-e^{\cos y} \sin y - x}$

Would you like further clarification on any of these steps?

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Related Questions:

1. How can implicit differentiation be applied to trigonometric functions?
2. What is the chain rule and how is it applied in implicit differentiation?
3. What are common applications of implicit differentiation in physics?
4. How do we handle exponential functions in implicit differentiation?
5. What steps can simplify solving for $\frac{dy}{dx}$ in implicit differentiation?

Tip: When using implicit differentiation, remember to apply the chain rule carefully, especially when dealing with compositions of functions like $e^{\cos y}$.