2. find the first derivative of f(x) = (4x² - 3x + 1)/√x

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\frac{4x^{2}-3x + 1}{\sqrt{x}}$ as $f(x)=4x^{\frac{3}{2}}-3x^{\frac{1}{2}}+x^{-\frac{1}{2}}$ using the rule $\frac{a + b + c}{d}=\frac{a}{d}+\frac{b}{d}+\frac{c}{d}$ and $\frac{x^{n}}{\sqrt{x}}=x^{n-\frac{1}{2}}$.

Step2: Apply the power - rule for differentiation

The power - rule states that if $y = x^{n}$, then $y^\prime=nx^{n - 1}$. For $y_1 = 4x^{\frac{3}{2}}$, $y_1^\prime=4\times\frac{3}{2}x^{\frac{3}{2}-1}=6x^{\frac{1}{2}}$. For $y_2=-3x^{\frac{1}{2}}$, $y_2^\prime=-3\times\frac{1}{2}x^{\frac{1}{2}-1}=-\frac{3}{2}x^{-\frac{1}{2}}$. For $y_3 = x^{-\frac{1}{2}}$, $y_3^\prime=-\frac{1}{2}x^{-\frac{1}{2}-1}=-\frac{1}{2}x^{-\frac{3}{2}}$.

Step3: Find the derivative of the function

$f^\prime(x)=y_1^\prime + y_2^\prime+y_3^\prime=6x^{\frac{1}{2}}-\frac{3}{2}x^{-\frac{1}{2}}-\frac{1}{2}x^{-\frac{3}{2}}$.

Answer:

$6\sqrt{x}-\frac{3}{2\sqrt{x}}-\frac{1}{2x\sqrt{x}}$