To find $\frac{dy}{dx}$ using implicit differentiation, we will differentiate both sides of the equation with respect to $x$.

Given:

$\cos(xy) = \sin(x + y)$

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

1. Left-hand side: We apply the chain rule to differentiate $\cos(xy)$.
   • First, differentiate the outer function $\cos(u)$ where $u = xy$, yielding $-\sin(xy)$.
   • Now, differentiate the inner function $xy$ using the product rule.

$\frac{d}{dx}[\cos(xy)] = -\sin(xy) \cdot \frac{d}{dx}(xy)$

The derivative of $xy$ with respect to $x$ is:

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

So, the derivative of the left-hand side is:

$-\sin(xy) \cdot (y + x\frac{dy}{dx})$

2. Right-hand side: Differentiate $\sin(x + y)$ using the chain rule.
   • The derivative of $\sin(u)$ where $u = x + y$ is $\cos(x + y)$.
   • The derivative of $x + y$ is $1 + \frac{dy}{dx}$.

Thus, the derivative of the right-hand side is:

$\cos(x + y) \cdot (1 + \frac{dy}{dx})$

Step 2: Set up the equation.

Now, we have:

$-\sin(xy) \cdot (y + x\frac{dy}{dx}) = \cos(x + y) \cdot (1 + \frac{dy}{dx})$

Step 3: Solve for $\frac{dy}{dx}$.

Expand both sides:

$-\sin(xy) \cdot y - \sin(xy) \cdot x \frac{dy}{dx} = \cos(x + y) \cdot 1 + \cos(x + y) \cdot \frac{dy}{dx}$

Now, collect terms involving $\frac{dy}{dx}$ on one side and the rest on the other:

$-\sin(xy) \cdot x \frac{dy}{dx} - \cos(x + y) \cdot \frac{dy}{dx} = \cos(x + y) + \sin(xy) \cdot y$

Factor out $\frac{dy}{dx}$ from the left-hand side:

$\frac{dy}{dx} \left( -\sin(xy) \cdot x - \cos(x + y) \right) = \cos(x + y) + \sin(xy) \cdot y$

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

$\frac{dy}{dx} = \frac{\cos(x + y) + \sin(xy) \cdot y}{-\sin(xy) \cdot x - \cos(x + y)}$

This is the derivative $\frac{dy}{dx}$ using implicit differentiation.

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Would you like more details on any part of this? Or have any questions?

Here are some related questions you might find useful:

1. How do we apply the chain rule in implicit differentiation?
2. What is the product rule in differentiation, and how is it used here?
3. How can we simplify the final expression for $\frac{dy}{dx}$?
4. What is the significance of implicit differentiation in solving equations?
5. How would the solution change if the equation involved only explicit functions of $x$ and $y$?

Tip: When performing implicit differentiation, always remember to treat $y$ as a function of $x$, meaning that any derivative of $y$ with respect to $x$ (such as $\frac{dy}{dx}$) must be included.