Answer:
[tex]f^{\prime}(x)=\frac{x^2-3}{x^2}[/tex]
Explanation:
Given the function f(x) defined as follows:
[tex]f(x)=\frac{x^2+3}{x}[/tex]
We use the quotient rule to differentiate this.
[tex]\text{Quotient Rule:}\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}[/tex]
From f(x):
[tex]\begin{gathered} u=x^2+3,\frac{du}{dx}=2x \\ v=x,\frac{dv}{dx}=1 \end{gathered}[/tex]
Therefore:
[tex]\begin{gathered} f^{\prime}(x)=\frac{x(2x)-(x^2+3)}{x^2} \\ =\frac{2x^2-x^2-3}{x^2} \\ f^{\prime}(x)=\frac{x^2-3}{x^2} \end{gathered}[/tex]
