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

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\frac{4x^{2}-3x + 1}{\sqrt[4]{x}}$ as $f(x)=4x^{2}\cdot x^{-\frac{1}{4}}-3x\cdot x^{-\frac{1}{4}}+x^{-\frac{1}{4}}=4x^{\frac{7}{4}}-3x^{\frac{3}{4}}+x^{-\frac{1}{4}}$ using the rule $\frac{a^{m}}{a^{n}}=a^{m - n}$.

Step2: Apply the power - rule for differentiation

The power - rule states that if $y = x^{n}$, then $y^\prime=nx^{n - 1}$. For $y = 4x^{\frac{7}{4}}$, $y^\prime=4\times\frac{7}{4}x^{\frac{7}{4}-1}=7x^{\frac{3}{4}}$. For $y=-3x^{\frac{3}{4}}$, $y^\prime=-3\times\frac{3}{4}x^{\frac{3}{4}-1}=-\frac{9}{4}x^{-\frac{1}{4}}$. For $y = x^{-\frac{1}{4}}$, $y^\prime=-\frac{1}{4}x^{-\frac{1}{4}-1}=-\frac{1}{4}x^{-\frac{5}{4}}$.

Step3: Combine the derivatives

$f^\prime(x)=7x^{\frac{3}{4}}-\frac{9}{4}x^{-\frac{1}{4}}-\frac{1}{4}x^{-\frac{5}{4}}$

Answer:

$f^\prime(x)=7x^{\frac{3}{4}}-\frac{9}{4x^{\frac{1}{4}}}-\frac{1}{4x^{\frac{5}{4}}}$