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The power rule is a fundamental concept in calculus that simplifies taking derivatives of functions written as exponentials. It's very handy because it allows you to quickly find the rate of change of functions. Essentially, the rule states: for a function of the form \(y = ax^n\), the derivative \(y'\) is \(anx^{n-1}\).
In the context of our example, we have the function \(y(t) = Ct^3\). Applying the power rule means multiplying the exponent (3) by the coefficient (C) and subtracting 1 from the exponent to get \(y'(t) = 3Ct^2\).
The steps are straightforward, and it's a tool you'll use often in calculus.

• Identify the exponent \(n\).
• Multiply \(n\) by the coefficient.
• Subtract 1 from the exponent for \(x^{n-1}\).

Remember, these steps allow you to find the derivative quickly and accurately.

A derivative represents the rate of change or the slope of a function at a given point. When you hear "derivative," think "how is something changing?" In the world of calculus, it's a powerful tool for understanding dynamics.
For the function \(y(t) = Ct^3\), the derivative \(y'(t)\) tells us how \(y\) changes as \(t\) changes. Using the power rule, we calculated it to be \(y'(t) = 3Ct^2\).
Derivatives are used to predict the behavior of functions, like how fast a car is accelerating or how quickly a population is growing. In problems involving differential equations, derivatives help us connect rates of change with the original function.

• Calculate the derivative using rules like the power rule.
• Interpret the derivative as a rate of change.
• Use derivatives in equations to find solutions or verify existing ones.

Understanding derivatives helps you unlock many aspects of mathematical analysis.

Solution verification is a vital step in ensuring your results are correct in mathematics, especially in differential equations. This process involves confirming that the proposed solution truly satisfies the differential equation.
To verify the solution, we substituted \(y(t) = C t^3\) and \(y'(t) = 3Ct^2\) back into the differential equation \(ty'(t) - 3y(t) = 0\). After replacement, we simplify:
Substituting gives \(t(3Ct^2) - 3(Ct^3) = 0\), which simplifies to \(3Ct^3 - 3Ct^3 = 0\). The left side equals the right side, showing that the original function is indeed a solution.
This ensures:

• Your calculations are correct.
• The function behaves as required by the differential equation.
• Any constants introduced are appropriate.

This careful step is crucial in confirming your answers are valid.

Substitution in differential equations is a method used to test if a proposed solution satisfies the equation. It's all about inserting your function and its derivatives into the equation to see if it holds.
Here's how it worked in our scenario: We started with the differential equation \(ty'(t) - 3y(t) = 0\). Then, we substituted \(y(t) = Ct^3\) and \(y'(t) = 3Ct^2\) into this equation. After doing this, the expression became \(3Ct^3 - 3Ct^3\), which simplified to zero, matching the right side of the equation.
Substitution is straightforward but powerful, verifying the actual alignment between your differential equation and the supposed solution.

• Start with the original function.
• Find the necessary derivative(s).
• Insert these into the differential equation.
• Simplify to check if the terms equal as needed.

This method ensures that the math works out, proving or disproving whether you have the correct function as a solution.