## Calculus 10th Edition

The derivative is $2t+\frac{12}{t^4}$.
First, rewrite the function as $y = t^2-4t^{-3}$. To find y', differentiate each individual term using the power rule $((x^n)' = nx^{n-1})$. The derivative of $t^2$ is $2t$. The derivative of $-4t^{-3}$ is $(-4)(-3)(t^{-3-1})$ which simplifies to $\frac{12}{t^4}$. Adding the answers together, we get the derivative to be $2t+\frac{12}{t^4}$.