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The given mathematical expression is:

\[
y = t^3 \cdot \tan(4t)
\]

We are asked to differentiate this function with respect to \( t \).

This is a product of two functions: \( t^3 \) and \( \tan(4t) \), so we will apply the **product rule** for differentiation, which is:

\[
\frac{d}{dt}[u \cdot v] = u' \cdot v + u \cdot v'
\]

Let:
- \( u = t^3 \), with \( u' = 3t^2 \)
- \( v = \tan(4t) \)

To differentiate \( v = \tan(4t) \), we will use the **chain rule**. The derivative of \( \tan(x) \) is \( \sec^2(x) \), so:

\[
v' = \frac{d}{dt}[\tan(4t)] = 4 \sec^2(4t)
\]

Now applying the product rule:

\[
\frac{dy}{dt} = (3t^2) \cdot \tan(4t) + (t^3) \cdot (4 \sec^2(4t))
\]

Simplifying the expression:

\[
\frac{dy}{dt} = 3t^2 \tan(4t) + 4t^3 \sec^2(4t)
\]

This is the derivative of the given function.

---

Would you like more details, or do you have any questions? 

Here are 5 related questions for further practice:
1. How would the derivative change if the function were \( y = t^4 \tan(5t) \)?
2. What is the integral of \( t^2 \tan(4t) \) with respect to \( t \)?
3. How would you find the second derivative of \( y = t^3 \tan(4t) \)?
4. Can you apply the product rule to \( y = e^t \cdot \sin(3t) \) and find its derivative?
5. What is the derivative of \( y = t^3 \sec^2(4t) \)?

**Tip:** Whenever you encounter a product of functions, always check whether the product rule or chain rule applies.

함수 y = t^3 tan(4t)를 미분하시오.

Learn how to differentiate the function y = t^3 tan(4t) using the product and chain rule. This detailed solution walks through key calculus concepts like the product rule and the chain rule. Suitable for high school students.

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