91Ó°ÊÓ

The power rule is one of the most fundamental tools in calculus for taking derivatives. It helps us find the derivative of a function when the variable is raised to a power. The rule is quite straightforward: if you have a term like \(t^n\), its derivative is \(nt^{n-1}\).

This means you multiply the entire term by the original exponent and then subtract one from the exponent.

Let's look at an example from the exercise: for the term \(5t^3\), you would multiply \(5\) by \(3\) (the exponent) to get \(15\), and then reduce the exponent by one, resulting in \(15t^2\).

The same applies to \(-3t^5\); it becomes \(-15t^4\) after applying the power rule. This makes finding derivatives quicker and easier.

When dealing with derivatives, another rule that often comes in handy is the product rule. This rule is essential when we have two functions multiplied together.

The product rule states that if a function \(s(t)\) is made up of two functions, say \(u(t)\) and \(v(t)\), then the derivative is \(u'v + uv'\).

In the exercise, after finding the first derivative \(15t^2(1-t^2)\), we need the second derivative which is more complicated because we have a product of functions.

By identifying \(u = 15t^2\) and \(v = 1-t^2\), we first find \(u' = 30t\) and \(v' = -2t\) and then apply the product rule to combine them. This results in the complete derivative \((30t(1-t^2) + 15t^2(-2t))\).
This step ties the product rule with our simplifying tools to solve problems involving products of functions.

The first derivative is crucial in calculus. It represents the rate of change of a function – in simpler terms, how fast or slow something is changing at any given point.

In the exercise, the first derivative is found by applying the power rule to each term. Starting from \(s = 5t^3 - 3t^5\), using the power rule gives \(15t^2 - 15t^4\).
The process of simplification follows, factoring it to \(15t^2(1-t^2)\), condensing the expression into a simpler, more manageable form.

This form is not only simpler but useful when calculating higher-order derivatives as it makes the derivative explicit, allowing for easier further manipulation.

The second derivative of a function gives us insight into the concavity and the rate of change of the rate of change. To find it, we continue from where we left off with the first derivative.

In the exercise, the first derivative \(15t^2(1-t^2)\) is taken further by employing the product rule to get more detail on its rate of change.

Applying the rule provides \(30t(1-t^2) + 15t^2(-2t)\), simplifying each term to reach \(30t - 60t^3\).
This expression \(30t(1 - 2t^2)\) after factoring, tells us not only about the change itself but how the change accelerates or decelerates.

Understanding second derivatives is vital for grasping the bigger picture of a function's behavior, looking deeper into its changes over time or space.