91Ó°ÊÓ

The power rule is a fundamental concept in calculus differentiation, making it easier to handle functions where variables are raised to a power. The rule states: if you have a function of the form \(f(t) = t^n\), the derivative \(\frac{d}{dt} f(t)\) is \(nt^{n-1}\). This means you multiply by the power and then lower the power by one. For our problem, the function terms \(-2t^{-1}\) and \(4t^{-2}\) are in the form that suits the power rule perfectly because each term is a power of \(t\).
Steps to apply the power rule:\

• Identify each variable expression raised to a power.


• Take the exponent, multiply it by the coefficient of the term.


• Reduce the exponent by one, keeping the base the same.

Using this method, the power rule simplifies the process of finding derivatives and is a crucial tool when differentiating polynomial-like functions.

Finding the first derivative means determining the rate at which the original function changes. It involves applying the differentiation process to our rewritten function \(s = -2t^{-1} + 4t^{-2}\). Using the power rule here means:
The derivative \(\frac{d}{dt} (-2t^{-1})\) becomes \(2t^{-2}\), indicating how fast the term \(-2t^{-1}\) changes with respect to \(t\), showing it's inversely proportional to \(t^2\).
Similarly, \(\frac{d}{dt}(4t^{-2})\) results in \(-8t^{-3}\). This derivative indicates a faster change inversely proportional to \(t^3\) due to the higher power.
The combined result is the first derivative, \(\frac{ds}{dt} = 2t^{-2} - 8t^{-3}\), which represents the original function's overall rate of change.

The process of finding the second derivative is essentially differentiating the first derivative. It provides information about the curvature or concavity of the original function \(s = -2t^{-1} + 4t^{-2}\). To obtain this, take the first derivative \(\frac{ds}{dt} = 2t^{-2} - 8t^{-3}\) and differentiate it once more.

• For \(2t^{-2}\), differentiating gives \(-4t^{-3}\), showing a further decrease, indicating increasing rapidity of initial change at increasing \(t^3\).


• For \(-8t^{-3}\), differentiating provides \(24t^{-4}\), suggesting a much steeper increasing curve at increasing \(t^4\).

After finding these, the second derivative \(\frac{d^2s}{dt^2} = -4t^{-3} + 24t^{-4}\) offers insight into not just how the rate of change itself changes, but how the dynamics of the original function \(s\) change over time, particularly how it might accelerate or decelerate.