To find $\frac{dy}{dx}$ from the equation $x^3 + \sin(xy) = 3$, we will use implicit differentiation.

Steps:

1. Differentiate both sides of the equation with respect to $x$:

$\frac{d}{dx}(x^3 + \sin(xy)) = \frac{d}{dx}(3)$

The right-hand side, being a constant (3), has a derivative of 0:

$\frac{d}{dx}(3) = 0$

Now, let's differentiate the left-hand side.

2. Differentiate $x^3$:

$\frac{d}{dx}(x^3) = 3x^2$

3. Differentiate $\sin(xy)$:

We need to use the chain rule and product rule here. Consider $\sin(xy)$ as a composite function. First, differentiate the outer function (sin), then the inner function $xy$:

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

Now, apply the product rule to $xy$:

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

Thus, the derivative of $\sin(xy)$ is:

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

4. Putting it all together:

Now, we have:

$3x^2 + \cos(xy)(y + x \frac{dy}{dx}) = 0$

5. Solve for $\frac{dy}{dx}$:

Isolate the term involving $\frac{dy}{dx}$:

$\cos(xy)(y + x \frac{dy}{dx}) = -3x^2$

Now, expand the left-hand side:

$\cos(xy) y + \cos(xy) x \frac{dy}{dx} = -3x^2$

Next, isolate the term with $\frac{dy}{dx}$:

$\cos(xy) x \frac{dy}{dx} = -3x^2 - \cos(xy) y$

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

$\frac{dy}{dx} = \frac{-3x^2 - \cos(xy) y}{\cos(xy) x}$

Final answer:

$\frac{dy}{dx} = \frac{-3x^2 - \cos(xy) y}{\cos(xy) x}$

Would you like further clarification or details?

Related Questions:

1. What is implicit differentiation, and why do we use it in this case?
2. How does the product rule apply when differentiating $xy$?
3. Can we solve for $y$ explicitly from $x^3 + \sin(xy) = 3$?
4. How does the chain rule work when differentiating composite functions like $\sin(xy)$?
5. What would change if the equation were $x^3 + \cos(xy) = 3$?

Tip: Implicit differentiation is useful when dealing with equations where $y$ is a function of $x$ but is not easily isolated.