Answer
$f'(x)=1-\frac{8}{x^3}$ .
Work Step by Step
First, split the numerator:
$f(x) = \frac{x^3}{x^2}-\frac{3x^2}{x^2}+\frac{4}{x^2}=x-3+\frac{4}{x^2}$.
Rewrite the function as $f(x) = x-3+4x^{-2}$.
To differentiate, use the power rule $((x^n)'=nx^{n-1})$ on the individual terms:
The derivative of $x$ is $1$.
The derivative of $-3$ is $0$.
The derivative of $4x^{-2}$ is equal to $(4)(-2)(x^{-2-1})$ which simplifies to $-\frac{8}{x^3}$.
Summing everything together gives $f'(x)=1-\frac{8}{x^3}$.
