We are asked to implicitly differentiate the equation $x^4 + \sin(y) = x^3$. Here, both $x$ and $y$ are functions of each other, so we need to apply implicit differentiation. Let's proceed step by step.

Given:

$x^4 + \sin(y) = x^3$

Step 1: Differentiate each term with respect to $x$.

• The derivative of $x^4$ with respect to $x$ is: $\frac{d}{dx}(x^4) = 4x^3$

• The derivative of $\sin(y)$ with respect to $x$ requires the chain rule, as $y$ is implicitly a function of $x$: $\frac{d}{dx}(\sin(y)) = \cos(y) \cdot \frac{dy}{dx}$

• The derivative of $x^3$ with respect to $x$ is: $\frac{d}{dx}(x^3) = 3x^2$

Step 2: Put everything together.

Now substitute the derivatives into the original equation:

$4x^3 + \cos(y) \cdot \frac{dy}{dx} = 3x^2$

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

Rearrange the equation to isolate $\frac{dy}{dx}$:

$\cos(y) \cdot \frac{dy}{dx} = 3x^2 - 4x^3$

$\frac{dy}{dx} = \frac{3x^2 - 4x^3}{\cos(y)}$

Thus, the derivative $\frac{dy}{dx}$ is:

$\boxed{\frac{dy}{dx} = \frac{3x^2 - 4x^3}{\cos(y)}}$

Do you have any questions, or would you like further details?

5 Related Questions:

1. How would you differentiate implicitly if $y$ was a more complex function like $y = e^x$?
2. What happens if we differentiate the equation twice with respect to $x$?
3. How does implicit differentiation work when there are multiple functions of $y$?
4. How would the result change if we had $x^2 \sin(y)$ instead of $\sin(y)$?
5. Can we use implicit differentiation to find second derivatives? How would that work?

Tip: Implicit differentiation is particularly useful when you cannot easily solve for $y$ as a function of $x$ before differentiating.