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Let's solve each part one by one.

### (a) \( f(x) = x^2 e^x \)
We need to differentiate this using the **product rule**, since we have the product of two functions: \( x^2 \) and \( e^x \).

The product rule is:
\[
(fg)' = f'g + fg'
\]

Let:
- \( f_1(x) = x^2 \) and \( f_2(x) = e^x \)

Then:
- \( f_1'(x) = 2x \)
- \( f_2'(x) = e^x \)

Using the product rule:
\[
f'(x) = (2x)e^x + (x^2)(e^x) = 2x e^x + x^2 e^x
\]

### (b) \( g(u) = u^3(2u^2 + e^u - 1) \)

Here we again have a product of two functions, \( u^3 \) and \( (2u^2 + e^u - 1) \). So, we use the product rule again.

Let:
- \( g_1(u) = u^3 \)
- \( g_2(u) = 2u^2 + e^u - 1 \)

Now:
- \( g_1'(u) = 3u^2 \)
- \( g_2'(u) = 4u + e^u \)

Using the product rule:
\[
g'(u) = (3u^2)(2u^2 + e^u - 1) + (u^3)(4u + e^u)
\]

### (c) \( h(x) = \frac{4e^x - 1}{x^2 + 1} \)

Here, we need to apply the **quotient rule**, since this is a ratio of two functions. The quotient rule is:
\[
\left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2}
\]

Let:
- \( h_1(x) = 4e^x - 1 \)
- \( h_2(x) = x^2 + 1 \)

Now:
- \( h_1'(x) = 4e^x \)
- \( h_2'(x) = 2x \)

Using the quotient rule:
\[
h'(x) = \frac{(4e^x)(x^2 + 1) - (4e^x - 1)(2x)}{(x^2 + 1)^2}
\]

### (d) \( j(x) = \left( \frac{x^2 + 1}{x} \right) e^x \)

Here we have both a quotient and a product, so we'll start by simplifying the quotient, if possible, then apply the product rule.

First, simplify the expression:
\[
j(x) = \left( \frac{x^2 + 1}{x} \right) e^x = \left( x + \frac{1}{x} \right) e^x
\]

Now, let:
- \( j_1(x) = x + \frac{1}{x} \)
- \( j_2(x) = e^x \)

Now differentiate each:
- \( j_1'(x) = 1 - \frac{1}{x^2} \)
- \( j_2'(x) = e^x \)

Using the product rule:
\[
j'(x) = \left(1 - \frac{1}{x^2}\right) e^x + \left(x + \frac{1}{x}\right) e^x
\]

### (e) \( k(x) = 3x^2 - \frac{2}{x^{7}} \)

Here, we can use basic power rule differentiation.

The power rule states:
\[
\frac{d}{dx}(x^n) = nx^{n-1}
\]

For each term:
- \( \frac{d}{dx}(3x^2) = 6x \)
- \( \frac{d}{dx}\left(-\frac{2}{x^7}\right) = -2 \cdot (-7) x^{-8} = 14x^{-8} \)

So, the derivative is:
\[
k'(x) = 6x + 14x^{-8}
\]

---

Would you like further details or explanations on any step? 

Here are 5 follow-up questions you might consider:
1. How does the product rule help in differentiating composite functions?
2. When is the quotient rule necessary, and can it be avoided by simplifying functions first?
3. What is the geometric interpretation of derivatives of these functions?
4. Could implicit differentiation be applied in any of these problems?
5. How do higher-order derivatives behave for exponential and polynomial functions?

**Tip:** Always try to simplify the function before applying differentiation rules to reduce complexity.

Find the derivatives of each of the following functions:
(a) f(x) = x^2 * e^x
(b) g(u) = u^3(2u^2 + e^u - 1)
(c) h(x) = (4e^x - 1) / (x^2 + 1)
(d) j(x) = ((x^2 + 1) / x) * e^x
(e) k(x) = 3x^2 - 2 / x^7

Learn how to find derivatives of functions involving products and quotients, including exponential and power functions. This problem covers key calculus concepts like the product rule, quotient rule, and power rule, suitable for students in grades 11-12 or early college.

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