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find the derivative of the algebraic function. f(x) = \\frac{2x - 3}{\\…

Question

find the derivative of the algebraic function.
f(x) = \frac{2x - 3}{\sqrt{x}}
f(x) =

Explanation:

Step1: Rewrite the function

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

Step2: Apply the power - rule for differentiation

The power - rule states that if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$.
For $y_1 = 2x^{\frac{1}{2}}$, $y_1^\prime=2\times\frac{1}{2}x^{\frac{1}{2}-1}=x^{-\frac{1}{2}}$.
For $y_2=-3x^{-\frac{1}{2}}$, $y_2^\prime=-3\times(-\frac{1}{2})x^{-\frac{1}{2}-1}=\frac{3}{2}x^{-\frac{3}{2}}$.

Step3: Find the derivative of $f(x)$

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

Answer:

$\frac{2x + 3}{2x\sqrt{x}}$