https://file.aiquickdraw.com/mathbot/file/1727628746012ktqom3mj.jpg

Let's break down the function and its derivative in the image:

Given:
\[
g(x) = \frac{3x^6}{e^x - 1}
\]
We also see its derivative \( g'(x) \) given by:
\[
g'(x) = \frac{18x^5(e^x - 1) - e^x(3x^6)}{(e^x - 1)^2}
\]

This derivative uses the quotient rule, which for a function \( f(x) = \frac{u(x)}{v(x)} \), gives:
\[
f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}
\]
Here, \( u(x) = 3x^6 \) and \( v(x) = e^x - 1 \).

### Verifying the derivative step-by-step:

1. **Derivative of the numerator \( u(x) = 3x^6 \)**:
   \[
   u'(x) = 18x^5
   \]

2. **Derivative of the denominator \( v(x) = e^x - 1 \)**:
   \[
   v'(x) = e^x
   \]

3. **Applying the quotient rule**:
   \[
   g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}
   \]
   Substituting the values:
   \[
   g'(x) = \frac{18x^5(e^x - 1) - 3x^6e^x}{(e^x - 1)^2}
   \]

This matches the expression given in the image.

Would you like me to simplify the derivative further or help with another part of the problem?

Here are some related questions for further exploration:
1. What is the second derivative \( g''(x) \) of this function?
2. How does this function behave as \( x \to 0 \)?
3. What is the limit of \( g(x) \) as \( x \to \infty \)?
4. Can we find the critical points of \( g(x) \) by setting \( g'(x) = 0 \)?
5. What is the physical or practical application of this type of function?

**Tip:** Always apply the quotient rule carefully by differentiating both the numerator and denominator separately before combining.

Differentiate the function g(x) = 3x^6 / (e^x - 1).

Learn how to differentiate g(x) = 3x^6 / (e^x - 1) step-by-step using the quotient rule. This example demonstrates key calculus concepts such as the derivative of exponential functions and power functions, suitable for college-level or AP Calculus students.

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