3.) Let w(x, y) = sin x + f(sin y - sin x) , where f / \mathbb{R} -> \mathbb{R} is a smooth function. (Here, f(sin y - sin x) is the composition of and the expres- sion sin y - sin x ) (i) (4 pts) Write down the formulas for partial w partial x and partial w partial y . Your answers will involve . f'

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.

Let w(x, y) = sin x + f(sin y - sin x), where f : R -> R is a smooth function. (i) Write down the formulas for partial derivatives of w with respect to x and y. Your answers will involve f'.

Learn how to compute the partial derivatives of w(x, y) = sin x + f(sin y - sin x) using the chain rule. This problem explores key concepts in multivariable calculus and provides a detailed solution, perfect for undergraduate calculus students.

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