https://file.aiquickdraw.com/mathbot/file/172766401162496mmybc5.jpg

Let's analyze the problem in the image step by step.

### Given:
You are tasked to compute the first derivative \( y'(x) \) and the second derivative \( y''(x) \) for the function:

\[
y(x) = \frac{3e^x}{x}
\]

#### 1. First Derivative \( y'(x) \):

We need to differentiate the function using the **quotient rule**:

\[
y'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{v(x)^2}
\]
where \( u(x) = 3e^x \) and \( v(x) = x \).

- \( u'(x) = 3e^x \) (since the derivative of \( e^x \) is \( e^x \))
- \( v'(x) = 1 \) (the derivative of \( x \) is \( 1 \))

Applying the quotient rule:

\[
y'(x) = \frac{x(3e^x) - 3e^x(1)}{x^2}
\]

Simplify the numerator:

\[
y'(x) = \frac{3xe^x - 3e^x}{x^2}
\]

Factor out \( 3e^x \) from the numerator:

\[
y'(x) = \frac{3e^x(x - 1)}{x^2}
\]

This matches the correct answer for \( y'(x) \) given in the image.

#### 2. Second Derivative \( y''(x) \):

Now, we need to differentiate \( y'(x) = \frac{3e^x(x - 1)}{x^2} \) again. We'll apply the quotient rule again.

Let:
- \( u(x) = 3e^x(x - 1) \)
- \( v(x) = x^2 \)

First, calculate the derivatives:
- \( u'(x) = 3e^x[(x - 1)' + (x - 1)e^x] = 3e^x(1 + x - 1) = 3e^x \cdot x \)
- \( v'(x) = 2x \)

Now apply the quotient rule again:

\[
y''(x) = \frac{x^2(3e^x x) - 3e^x(x - 1)2x}{x^4}
\]

Simplify the numerator:

\[
y''(x) = \frac{3e^x x^3 - 6e^x x(x - 1)}{x^4}
\]

Simplify further:

\[
y''(x) = \frac{3e^x [x^3 - 2x(x - 1)]}{x^4}
\]

Finally, simplify the terms inside the bracket:

\[
y''(x) = \frac{3e^x [x^3 - 2x^2 + 2x]}{x^4}
\]

So, the second derivative is not simply \( \frac{3e^x}{x^4} \), which is indicated as the incorrect answer in the image. The error likely arises from missing terms due to the product rule inside the derivative of the numerator.

Would you like more detailed steps or further clarification on any part?

Here are 5 related questions that can extend your learning:
1. How would you find the third derivative for this function?
2. Can you apply the product rule and quotient rule in combination?
3. What is the interpretation of the first and second derivatives geometrically?
4. How would the process change if the function involved more complex exponents?
5. Can you graph the original function and its first and second derivatives?

**Tip:** When differentiating quotients, it's crucial to ensure both the numerator and denominator's derivatives are handled carefully, especially when applying the product rule inside the quotient rule.

Calculate the first and second derivatives of y(x) = 3e^x / x.

This problem explores how to calculate the first and second derivatives of the function y(x) = 3e^x / x using the quotient rule in calculus. Suitable for students in Grades 11-12, it includes detailed steps and explanations of the quotient rule application.

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