The instructions in the image ask to differentiate the functions and factor out where possible. Let’s go through each function, applying the product and chain rules for differentiation, and simplifying where appropriate.

Differentiation Process

1. Function (a): $f(x) = 2x \cdot e^{3x}$

   • Use the product rule: $f'(x) = (2x)' \cdot e^{3x} + 2x \cdot (e^{3x})'$.
   • Derivative: $f'(x) = 2 \cdot e^{3x} + 2x \cdot 3e^{3x} = (2 + 6x)e^{3x}$.

2. Function (b): $g(x) = (x - 3) \cdot e^{2x}$

   • Use the product rule: $g'(x) = (x - 3)' \cdot e^{2x} + (x - 3) \cdot (e^{2x})'$.
   • Derivative: $g'(x) = e^{2x} + (x - 3) \cdot 2e^{2x} = (1 + 2(x - 3))e^{2x} = (2x - 5)e^{2x}$.

3. Function (c): $f(t) = t^2 \cdot e^{\frac{1}{2}t} + 5$

   • Use the product rule for the first part and the constant rule for the second.
   • Derivative: $f'(t) = (t^2)' \cdot e^{\frac{1}{2}t} + t^2 \cdot \left(e^{\frac{1}{2}t}\right)'$.
   • Derivative calculation: $f'(t) = 2t \cdot e^{\frac{1}{2}t} + t^2 \cdot \frac{1}{2}e^{\frac{1}{2}t} = \left(2t + \frac{t^2}{2}\right)e^{\frac{1}{2}t}$.

4. Function (d): $g(t) = (t^3 - 2t^2) \cdot e^{-2t}$

   • Use the product rule: $g'(t) = (t^3 - 2t^2)' \cdot e^{-2t} + (t^3 - 2t^2) \cdot (e^{-2t})'$.
   • Derivative: $g'(t) = (3t^2 - 4t)e^{-2t} + (t^3 - 2t^2)(-2)e^{-2t}$.
   • Simplify: $g'(t) = (3t^2 - 4t - 2t^3 + 4t^2)e^{-2t} = (-2t^3 + 7t^2 - 4t)e^{-2t}$.

5. Function (e): $f(x) = (2x - 4) \cdot e^{0.5x}$

   • Use the product rule: $f'(x) = (2x - 4)' \cdot e^{0.5x} + (2x - 4) \cdot (e^{0.5x})'$.
   • Derivative: $f'(x) = 2 \cdot e^{0.5x} + (2x - 4) \cdot 0.5e^{0.5x}$.
   • Simplify: $f'(x) = \left(2 + (x - 2)\right)e^{0.5x} = (x)e^{0.5x}$.

6. Function (f): $h(x) = (x^2 - x) \cdot e^{-0.01x}$

   • Use the product rule: $h'(x) = (x^2 - x)' \cdot e^{-0.01x} + (x^2 - x) \cdot (e^{-0.01x})'$.
   • Derivative: $h'(x) = (2x - 1)e^{-0.01x} + (x^2 - x)(-0.01)e^{-0.01x}$.
   • Simplify: $h'(x) = (2x - 1 - 0.01x^2 + 0.01x)e^{-0.01x} = (-0.01x^2 + 2.01x - 1)e^{-0.01x}$.

Would you like further details or clarification on any specific step?

Here are some related questions for further exploration:

1. How does the product rule work in different types of functions?
2. What are some applications of exponential functions in real-world problems?
3. How can factoring simplify complex derivatives?
4. What role does the chain rule play when differentiating composed functions?
5. How does the choice of base (e.g., $e$ vs. other constants) affect differentiation?

Tip: When differentiating exponential functions, remember that the derivative of $e^{u(x)}$ is $u'(x) \cdot e^{u(x)}$.