Differentiate \( f(x)=\frac{x^{2}-4 x+4}{x^{2}} \) Answer: \( \frac{d f}{d x}=\square \)

To differentiate the function \( f(x) = \frac{x^{2}-4x+4}{x^{2}} \), we first simplify it. Notice that the numerator can be factored as \( (x-2)^{2} \), so we have \( f(x) = \frac{(x-2)^{2}}{x^{2}} \). This can be rewritten as \( f(x) = \left(\frac{x-2}{x}\right)^{2} \). Now, we can differentiate it using the quotient rule or the chain rule. Applying the quotient rule: \[ \frac{d f}{d x} = \frac{(x^{2}) \cdot (2(x-2)(\frac{d}{dx}(x-2)))-(x^{2}-4x+4)(2x)}{(x^{2})^{2}} \] Calculating this gives us: \[ \frac{d f}{d x} = \frac{(2(x^{2})(x-2) - 2x(x^{2}-4x+4))}{x^{4}} \] So, after simplification, we find that: \[ \frac{d f}{d x} = \frac{2(x^{2}-4x)}{x^{4}} = \frac{2(x(x-4))}{x^{4}} = \frac{2}{x^{3}}(x-4) \] Thus, the final derivative is: \[ \frac{d f}{d x} = \frac{2(x-4)}{x^{3}} \]