Answer:
[tex]\boxed{f^{\prime}(x)=x^2^{}-\frac{1}{2}}[/tex]Explanation:
Step 1. The function we have is:
[tex]f(x)=\frac{x^3}{3}-\frac{x}{2}[/tex]And we are asked to find the derivative of the function. The rule to find the derivative for this type of function is:
[tex]\begin{gathered} \text{for a function }of\text{ the form} \\ f(x)=ax^n \\ \text{The derivative is:} \\ f^{\prime}(x)=a(n)x^{n-1} \end{gathered}[/tex]Step 2. Before we apply the derivative rule, remember the following:
[tex]\begin{gathered} \text{for a function } \\ f(x)=g(x)+h(x) \\ \text{The derivative is:} \\ f^{\prime}(x)=g^{\prime}(x)+h^{\prime}(x) \end{gathered}[/tex]This means that we need to derivate each part or term of the function and combine them for the total derivative.
Step 3. Apply the derivative rule from step 1 to the given function.
First we rewrite the function as follows:
[tex]\begin{gathered} f(x)=\frac{x^3}{3}-\frac{x}{2} \\ \downarrow \\ f(x)=\frac{1}{3}x^3-\frac{1}{2}x^1 \end{gathered}[/tex]Apply the derivative rule:
[tex]f^{\prime}(x)=\frac{1}{3}(3)x^{3-1}-\frac{1}{2}(1)x^{1-1}[/tex]Step 4. The last step is to simplify the expression:
[tex]\begin{gathered} f^{\prime}(x)=\frac{1}{3}(3)x^{3-1}-\frac{1}{2}(1)x^{1-1} \\ \downarrow \\ f^{\prime}(x)=\frac{1}{3}(3)x^2-\frac{1}{2}(1)x^0 \\ f^{\prime}(x)=x^2-\frac{1}{2}x^{^0} \\ \sin ce^{} \\ x^0=1 \\ \downarrow\text{ The result is }\downarrow \\ f^{\prime}(x)=x^2-\frac{1}{2} \end{gathered}[/tex]Answer:
[tex]\boxed{f^{\prime}(x)=x^2^{}-\frac{1}{2}}[/tex]