The given function is \( Y = (3x)^3 \sin(3x) \). Let's simplify and work on it.

### Simplified form:
\[
Y = 27x^3 \sin(3x)
\]

Now, if you're looking for derivatives, I'll compute the first derivative \( \frac{dY}{dx} \).

### First derivative (\( \frac{dY}{dx} \)):
We need to apply the product rule for differentiation. The product rule states that for two functions \( f(x) \) and \( g(x) \), the derivative of their product is:
\[
\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)
\]
Here:
- \( f(x) = 27x^3 \)
- \( g(x) = \sin(3x) \)

First, let's differentiate each part:
1. \( f'(x) = \frac{d}{dx}[27x^3] = 81x^2 \)
2. \( g'(x) = \frac{d}{dx}[\sin(3x)] = 3 \cos(3x) \)

Now, apply the product rule:
\[
\frac{dY}{dx} = 81x^2 \sin(3x) + 27x^3 \cdot 3 \cos(3x)
\]
Simplifying the second term:
\[
\frac{dY}{dx} = 81x^2 \sin(3x) + 81x^3 \cos(3x)
\]

So, the first derivative of the function is:
\[
\frac{dY}{dx} = 81x^2 \sin(3x) + 81x^3 \cos(3x)
\]

Would you like to explore more on this, or need clarification on any part?

### Related Questions:
1. How do you apply the product rule in calculus?
2. What is the second derivative of \( Y = 27x^3 \sin(3x) \)?
3. How does the chain rule work when differentiating \( \sin(3x) \)?
4. How do you find critical points for \( Y = 27x^3 \sin(3x) \)?
5. How would you compute the integral of \( Y = 27x^3 \sin(3x) \)?

### Tip:
When using the product rule, always double-check your individual derivatives before combining them, especially when dealing with complex functions like trigonometric ones.

Learn how to differentiate the function Y = (3x)^3 sin(3x) by applying the product rule. This solution covers key calculus concepts like the product rule and trigonometric differentiation. Suitable for students in Grades 11-12 or college-level calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation