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The power rule is a fundamental concept in differential calculus. It provides a straightforward way to find the derivative of a function that is a power of a variable. This rule states that if you have a function of the form \(y = x^n\), where \(n\) is a constant, the derivative of \(y\) with respect to \(x\) is \(\frac{d}{dx}(x^n) = nx^{n-1}\).

In our given exercise, the function is \(y = (x + 10)^{6}\). To find the first derivative, we apply the power rule directly to obtain \(\frac{dy}{dx} = 6(x + 10)^{5}\). The power rule is useful because it simplifies finding derivatives of polynomial functions. Its simplicity and consistency make it a preferred method when faced with powers of variables. Remember to always multiply the given power by the original expression raised to one power less.

In calculus, higher-order derivatives refer to the process of differentiating a function multiple times. The first derivative represents the rate of change, like velocity in physics. However, higher-order derivatives can indicate further rates of change, like acceleration.

For the given function \(y = (x + 10)^{6}\), we start with the first derivative: \(\frac{dy}{dx} = 6(x + 10)^{5}\). We then find the second derivative by applying the power rule again to get \(\frac{d^2 y}{dx^2} = 30(x + 10)^{4}\). This process can be continued to find the third derivative: \(\frac{d^3 y}{dx^3} = 120(x + 10)^{3}\).

The notation \(\frac{d^n y}{dx^n}\) is used to indicate \(n\)-th order derivatives. Higher-order derivatives are valuable in various fields, such as physics and engineering, for understanding dynamics.

Once we have the derivative, we can evaluate it at a specific point to find the rate of change at that location. This process involves substituting a particular \(x\)-value into the derivative equation.

In the exercise, we needed to find the third derivative of \(y = (x + 10)^{6}\) evaluated at \(x = 2\). We obtained the third derivative: \(\frac{d^3 y}{dx^3} = 120(x + 10)^{3}\). By substituting \(x = 2\) into this expression, we computed \(120(2 + 10)^3 = 207,360\).

This step is crucial because it gives the exact rate of change at a specific \(x\) value, which can be particularly useful in practical applications. In real-world scenarios, evaluating derivatives at certain points can provide insights into the behavior of physical systems or changes occurring at precise moments.