## Calculus 10th Edition

$f'(x)=1-\frac{8}{x^3}$ .
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 $f'(x)=1-\frac{8}{x^3}$.