To differentiate the given functions using the first principle, we will follow these steps:

First Principle of Differentiation

The derivative of a function $f(x)$ using the first principle is given by:

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

Now, let's apply this to each of the given functions.

2.1.1 $f(x) = \frac{1}{x}$

$f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}$

Find a common denominator:

$f'(x) = \lim_{h \to 0} \frac{\frac{x - (x+h)}{x(x+h)}}{h} = \lim_{h \to 0} \frac{-h}{h \cdot x(x+h)} = \lim_{h \to 0} \frac{-1}{x(x+h)}$

As $h \to 0$:

$f'(x) = \frac{-1}{x^2}$

2.1.2 $f(x) = \frac{1}{x^2}$

$f'(x) = \lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}$

Simplify the expression:

= \lim_{h \to 0} \frac{-2xh - h^2}{h \cdot x^2(x+h)^2} = \lim_{h \to 0} \frac{-2x - h}{x^2(x+h)^2}$$ As $$h \to 0$$: $$f'(x) = \frac{-2}{x^3}$$ ### 2.1.3 $$f(x) = 2x^2$$ $$f'(x) = \lim_{h \to 0} \frac{2(x+h)^2 - 2x^2}{h}$$ Expand $$(x+h)^2$$ and simplify: $$f'(x) = \lim_{h \to 0} \frac{2(x^2 + 2xh + h^2) - 2x^2}{h} = \lim_{h \to 0} \frac{4xh + 2h^2}{h}$$ Factor out $$h$$: $$f'(x) = \lim_{h \to 0} (4x + 2h)$$ As $$h \to 0$$: $$f'(x) = 4x$$ So, the derivatives are: 1. $$f'(x) = -\frac{1}{x^2}$$ for $$f(x) = \frac{1}{x}$$ 2. $$f'(x) = -\frac{2}{x^3}$$ for $$f(x) = \frac{1}{x^2}$$ 3. $$f'(x) = 4x$$ for $$f(x) = 2x^2$$ Would you like any further details or clarifications? Here are 5 related questions: 1. What is the first principle of differentiation, and why is it important? 2. How do we differentiate a function involving a power of $$x$$ using the first principle? 3. What is the difference between the first principle method and standard differentiation techniques? 4. Can the first principle method be used for differentiating trigonometric functions? 5. How can we apply the first principle to find the derivative of $$f(x) = \sqrt{x}$$? **Tip:** Always simplify the function as much as possible before applying the limit in the first principle of differentiation to make the process easier.