The problem in the image asks to "Find the derivative of $f(x) = x^2 + 2x$ using the first principle."

To solve this, we use the first principle of differentiation, which defines the derivative $f'(x)$ as:

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Step-by-step solution:

Given $f(x) = x^2 + 2x$, we follow these steps:

1. Find $f(x+h)$: $f(x+h) = (x+h)^2 + 2(x+h)$ Expanding this: $f(x+h) = x^2 + 2xh + h^2 + 2x + 2h$

2. Find $f(x+h) - f(x)$: Subtract $f(x) = x^2 + 2x$ from $f(x+h)$: $f(x+h) - f(x) = (x^2 + 2xh + h^2 + 2x + 2h) - (x^2 + 2x)$ Simplifying: $f(x+h) - f(x) = 2xh + h^2 + 2h$

3. Divide by $h$: $\frac{f(x+h) - f(x)}{h} = \frac{2xh + h^2 + 2h}{h}$ Simplifying further: $\frac{f(x+h) - f(x)}{h} = 2x + h + 2$

4. Take the limit as $h \to 0$: $f'(x) = \lim_{h \to 0} (2x + h + 2) = 2x + 2$

Thus, the derivative of $f(x) = x^2 + 2x$ using the first principle is:

$f'(x) = 2x + 2$

Would you like any further clarification on this solution? Here are five related questions to expand your understanding:

1. How is the first principle of differentiation related to the concept of limits?
2. What happens if we try to differentiate a constant function using the first principle?
3. Can the first principle be used to differentiate higher-order functions like cubic or quartic polynomials?
4. What role does the expansion of $f(x+h)$ play in applying the first principle?
5. How does using the first principle differ from using standard differentiation rules?

Tip: When applying the first principle, always expand and simplify expressions carefully before dividing by $h$.