A) - fist primeiple find \( y=\frac{1}{x^{2}} \)

To find the first principle (or the definition of the derivative) of the function \( y = \frac{1}{x^2} \), we will use the limit definition of the derivative. The derivative of a function \( f(x) \) at a point \( x \) is given by: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] In this case, our function is \( f(x) = \frac{1}{x^2} \). We will compute \( f'(x) \) using the limit definition. 1. **Calculate \( f(x+h) \)**: \[ f(x+h) = \frac{1}{(x+h)^2} \] 2. **Set up the limit**: \[ f'(x) = \lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} \] 3. **Combine the fractions in the numerator**: \[ \frac{1}{(x+h)^2} - \frac{1}{x^2} = \frac{x^2 - (x+h)^2}{x^2 (x+h)^2} \] Expanding \( (x+h)^2 \): \[ (x+h)^2 = x^2 + 2xh + h^2 \] Thus, \[ x^2 - (x+h)^2 = x^2 - (x^2 + 2xh + h^2) = -2xh - h^2 \] Therefore, \[ \frac{1}{(x+h)^2} - \frac{1}{x^2} = \frac{-2xh - h^2}{x^2 (x+h)^2} \] 4. **Substituting back into the limit**: \[ f'(x) = \lim_{h \to 0} \frac{-2xh - h^2}{h \cdot x^2 (x+h)^2} \] Simplifying the fraction: \[ f'(x) = \lim_{h \to 0} \frac{-2x - h}{x^2 (x+h)^2} \] 5. **Taking the limit as \( h \to 0 \)**: \[ f'(x) = \frac{-2x}{x^2 \cdot x^2} = \frac{-2x}{x^4} = \frac{-2}{x^3} \] Thus, the derivative of \( y = \frac{1}{x^2} \) is: \[ \boxed{f'(x) = -\frac{2}{x^3}} \]