This problem requires calculating the derivative of each given function $f(x)$ using the definition of the derivative:

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

Let's go through each case.

(a) $f(x) = x^2$

1. Substitute $f(x+h) = (x+h)^2$ and $f(x) = x^2$.

   $\frac{f(x+h) - f(x)}{h} = \frac{(x+h)^2 - x^2}{h}$

2. Expand $(x+h)^2 = x^2 + 2xh + h^2$:

   $= \frac{x^2 + 2xh + h^2 - x^2}{h} = \frac{2xh + h^2}{h} = \frac{h(2x + h)}{h} = 2x + h.$

3. Take the limit as $h \to 0$:

   $\lim_{h \to 0} (2x + h) = 2x.$

So, $f'(x) = 2x$.

(b) $f(x) = 5$

1. Since $f(x+h) = 5$ and $f(x) = 5$, the expression becomes:

   $\frac{f(x+h) - f(x)}{h} = \frac{5 - 5}{h} = \frac{0}{h} = 0.$

2. The limit as $h \to 0$ is simply 0.

So, $f'(x) = 0$.

(c) $f(x) = 2x^2 + x$

1. Substitute $f(x+h) = 2(x+h)^2 + (x+h)$ and $f(x) = 2x^2 + x$:

   $\frac{f(x+h) - f(x)}{h} = \frac{2(x+h)^2 + (x+h) - (2x^2 + x)}{h}$

2. Expand $2(x+h)^2 = 2(x^2 + 2xh + h^2) = 2x^2 + 4xh + 2h^2$ and $(x+h) = x + h$:

   $= \frac{(2x^2 + 4xh + 2h^2 + x + h) - (2x^2 + x)}{h}$

3. Simplify terms:

   $= \frac{4xh + 2h^2 + h}{h} = \frac{h(4x + 2h + 1)}{h} = 4x + 2h + 1.$

4. Take the limit as $h \to 0$:

   $\lim_{h \to 0} (4x + 2h + 1) = 4x + 1.$

So, $f'(x) = 4x + 1$.

Summary of Results

• (a) $f(x) = x^2 \Rightarrow f'(x) = 2x$
• (b) $f(x) = 5 \Rightarrow f'(x) = 0$
• (c) $f(x) = 2x^2 + x \Rightarrow f'(x) = 4x + 1$

Would you like further details on any of these steps, or do you have any questions?

Related Questions

1. What is the derivative of $f(x) = 3x^2 + 2x + 1$ using the same limit definition?
2. How do derivatives represent the slope of a function at a given point?
3. What is the physical interpretation of the derivative in terms of velocity, if $f(x)$ represents distance?
4. How does the definition of the derivative apply to trigonometric functions, such as $f(x) = \sin(x)$?
5. Can you explain how the derivative of a constant function is always zero?

Tip

When applying the limit definition of the derivative, simplifying algebraic expressions is key before taking the limit as $h \to 0$.