We are given a function $w(x, y) = \sin x + f(\sin y - \sin x)$, where $f: \mathbb{R} \to \mathbb{R}$ is a smooth function and the term $f(\sin y - \sin x)$ represents the composition of $f$ and the expression $\sin y - \sin x$.

We need to compute the partial derivatives $\frac{\partial w}{\partial x}$ and $\frac{\partial w}{\partial y}$.

Step 1: Compute $\frac{\partial w}{\partial x}$

We start by differentiating $w(x, y)$ with respect to $x$:

$w(x, y) = \sin x + f(\sin y - \sin x)$

The derivative of $\sin x$ with respect to $x$ is $\cos x$. Now, for the second term $f(\sin y - \sin x)$, we need to apply the chain rule. Let $u = \sin y - \sin x$, then:

$\frac{\partial}{\partial x} f(\sin y - \sin x) = f'(u) \cdot \frac{\partial}{\partial x} (\sin y - \sin x)$

Since $\sin y$ is independent of $x$, its derivative is zero, and $\frac{\partial}{\partial x} (-\sin x) = -\cos x$. Therefore:

$\frac{\partial}{\partial x} f(\sin y - \sin x) = f'(\sin y - \sin x) \cdot (-\cos x)$

Thus, the total derivative of $w$ with respect to $x$ is:

$\frac{\partial w}{\partial x} = \cos x - f'(\sin y - \sin x) \cdot \cos x$

Step 2: Compute $\frac{\partial w}{\partial y}$

Now, we differentiate $w(x, y)$ with respect to $y$:

$w(x, y) = \sin x + f(\sin y - \sin x)$

The derivative of $\sin x$ with respect to $y$ is zero since $\sin x$ is independent of $y$. For the term $f(\sin y - \sin x)$, again applying the chain rule:

$\frac{\partial}{\partial y} f(\sin y - \sin x) = f'(u) \cdot \frac{\partial}{\partial y} (\sin y - \sin x)$

The derivative of $\sin y$ with respect to $y$ is $\cos y$, and the derivative of $-\sin x$ with respect to $y$ is zero. Therefore:

$\frac{\partial}{\partial y} f(\sin y - \sin x) = f'(\sin y - \sin x) \cdot \cos y$

Thus, the total derivative of $w$ with respect to $y$ is:

$\frac{\partial w}{\partial y} = f'(\sin y - \sin x) \cdot \cos y$

Final Answer

• $\frac{\partial w}{\partial x} = \cos x - f'(\sin y - \sin x) \cdot \cos x$
• $\frac{\partial w}{\partial y} = f'(\sin y - \sin x) \cdot \cos y$

Would you like more details on the chain rule or any other part?

Here are 5 related questions for further practice:

1. What happens to $\frac{\partial w}{\partial x}$ and $\frac{\partial w}{\partial y}$ if $f(t) = t^2$?
2. How would you compute the second partial derivatives $\frac{\partial^2 w}{\partial x^2}$ and $\frac{\partial^2 w}{\partial y^2}$?
3. What are the conditions for $w(x, y)$ to be independent of $y$?
4. If $f(t) = e^t$, what do the partial derivatives become?
5. How do the partial derivatives change if $w(x, y) = \cos x + f(\cos y - \cos x)$?

Tip: Always apply the chain rule carefully when differentiating composite functions.