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We are tasked with finding \(\frac{dw}{dt}\) using the chain rule, where:

\[
w = \sin(8x) \cos(3y), \quad x = \frac{t}{4}, \quad y = t^3.
\]

### Step-by-Step Solution:

#### 1. **Chain Rule Setup:**
We know that:
\[
\frac{dw}{dt} = \frac{\partial w}{\partial x} \frac{dx}{dt} + \frac{\partial w}{\partial y} \frac{dy}{dt}.
\]
We need to compute \(\frac{\partial w}{\partial x}\), \(\frac{\partial w}{\partial y}\), \(\frac{dx}{dt}\), and \(\frac{dy}{dt}\).

#### 2. **Partial Derivatives of \(w\):**

- \(\frac{\partial w}{\partial x}\): Differentiating \(w = \sin(8x) \cos(3y)\) with respect to \(x\):
  \[
  \frac{\partial w}{\partial x} = 8 \cos(8x) \cos(3y) \quad \text{(using chain rule on \(\sin(8x)\))}.
  \]

- \(\frac{\partial w}{\partial y}\): Differentiating \(w = \sin(8x) \cos(3y)\) with respect to \(y\):
  \[
  \frac{\partial w}{\partial y} = -3 \sin(8x) \sin(3y) \quad \text{(using chain rule on \(\cos(3y)\))}.
  \]

#### 3. **Derivatives of \(x\) and \(y\) with respect to \(t\):**

- \(\frac{dx}{dt}\): Differentiating \(x = \frac{t}{4}\) with respect to \(t\):
  \[
  \frac{dx}{dt} = \frac{1}{4}.
  \]

- \(\frac{dy}{dt}\): Differentiating \(y = t^3\) with respect to \(t\):
  \[
  \frac{dy}{dt} = 3t^2.
  \]

#### 4. **Substituting into Chain Rule:**
Now substitute these into the chain rule expression:

\[
\frac{dw}{dt} = \left( 8 \cos(8x) \cos(3y) \right) \cdot \frac{1}{4} + \left( -3 \sin(8x) \sin(3y) \right) \cdot 3t^2.
\]

Simplifying:
\[
\frac{dw}{dt} = 2 \cos(8x) \cos(3y) - 9t^2 \sin(8x) \sin(3y).
\]

#### 5. **Substitute \(x = \frac{t}{4}\) and \(y = t^3\):**
Finally, substitute the expressions for \(x\) and \(y\) in terms of \(t\):
\[
x = \frac{t}{4}, \quad y = t^3,
\]
giving:
\[
\frac{dw}{dt} = 2 \cos\left( 8 \cdot \frac{t}{4} \right) \cos(3t^3) - 9t^2 \sin\left( 8 \cdot \frac{t}{4} \right) \sin(3t^3),
\]
which simplifies further to:
\[
\frac{dw}{dt} = 2 \cos(2t) \cos(3t^3) - 9t^2 \sin(2t) \sin(3t^3).
\]

This is the derivative \(\frac{dw}{dt}\) in terms of \(t\).

Would you like further details or clarification on any step?

Here are 5 related questions:
1. How does the chain rule apply to multivariable functions?
2. What happens if \(x\) and \(y\) are defined implicitly rather than explicitly?
3. Can we apply this process to other trigonometric or exponential functions?
4. What is the interpretation of the second derivative in this context?
5. How does the result change if we modify \(x(t)\) or \(y(t)\)?

**Tip:** Always verify partial derivatives with respect to each variable before applying the chain rule to ensure accuracy in multivariable calculus.

Use the Chain Rule to find dw/dt, where w = sin(8x) cos(3y), x = t/4, and y = t^3.

Learn how to apply the chain rule to find dw/dt where w = sin(8x) cos(3y), x = t/4, and y = t^3. This problem involves trigonometric functions and the chain rule in multivariable calculus, suitable for undergraduate calculus students.

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