Question

1. differentiate $f(x)=\frac{x^{2}-4x + 1}{x^{2}}$. answer: $\frac{df}{dx}=$

Explanation:

Step1: Simplify the function

First, rewrite $f(x)=\frac{x^{2}-4x + 1}{x^{2}}$ as $f(x)=1-\frac{4}{x}+\frac{1}{x^{2}}=1 - 4x^{-1}+x^{-2}$.

Step2: Differentiate term - by - term

Using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, for the constant term 1, $\frac{d}{dx}(1)=0$; for $-4x^{-1}$, $\frac{d}{dx}(-4x^{-1})=-4\times(-1)x^{-2}=4x^{-2}$; for $x^{-2}$, $\frac{d}{dx}(x^{-2})=-2x^{-3}$.

Step3: Combine the results

$f^\prime(x)=\frac{df}{dx}=0 + 4x^{-2}-2x^{-3}=\frac{4}{x^{2}}-\frac{2}{x^{3}}$.

Answer:

$\frac{4}{x^{2}}-\frac{2}{x^{3}}$