91Ó°ÊÓ

The concept of the derivative is a cornerstone of calculus. It represents the rate of change of a function as its input changes. In simpler terms, it tells you how the function behaves as you change the values of the input variable. For our function, which is \(f(x) = \sqrt{x}\), the derivative provides insight into how the square root changes with differing values of \(x\).

To find the derivative of \(\sqrt{x}\), we use the chain rule. The chain rule is a method for computing the derivative of a composite function. For \(f(x) = \sqrt{x} = x^{1/2}\), the derivative \(f'(x)\) is calculated as \(\frac{1}{2\sqrt{x}}\). This derivative tells us that the slope of the curve \(\sqrt{x}\) decreases as \(x\) increases. In the exercise, we are particularly interested in the value of this derivative at a specific point, \(x=9\).

The Taylor Series is a way to represent a function as an infinite sum of terms based on the function's derivatives at a single point. This series is incredibly useful because it allows us to approximate complex functions using polynomials, which are much simpler to work with.

For a function \(f(x)\) at a point \(a\), the Taylor series is represented as:
\[ f(a) + f'(a)\frac{(x-a)}{1!} + f''(a)\frac{(x-a)^2}{2!} + \cdots\]

In the context of our exercise, we want to find the Taylor series for \(\sqrt{x}\) centered at \(x=9\). Here, \(a=9\), and our task is to calculate the function values and its consequent derivatives at \(x=9\). This allows us to construct the series up to the desired degree, which in this case is the 5th degree.

Polynomial approximation involves using polynomials to estimate the values of more complicated functions. This approach is useful because polynomials are easy to differentiate and integrate, making them excellent candidates for approximation.

In our exercise, a 5th degree Taylor polynomial is used to approximate \(\sqrt{x}\). By considering terms up to the 5th derivative, we gain a very close approximation of the actual function for values of \(x\) that are near the center point, \(x=9\).

This approximation is given by the formula:
\[ f(9) + f'(9)\frac{(x-9)}{1!} + f''(9)\frac{(x-9)^2}{2!} + f'''(9)\frac{(x-9)^3}{3!} + f''''(9)\frac{(x-9)^4}{4!} + f'''''(9)\frac{(x-9)^5}{5!}\]

By plugging in the derivative values, the polynomial serves as a local approximation to the function \(\sqrt{x}\). It provides a practical method to compute square roots of numbers close to 9 without using a calculator.

Higher-order derivatives are the successive derivatives of a function, representing the rate of change of the previous derivative. In essence, they provide deeper insights into the behavior of functions.

For example, the second derivative can often tell us about the concavity of a function and whether a given point is a maximum, minimum, or point of inflection. For the function \(\sqrt{x}\), we are calculating derivatives up to the 5th order:

• The first derivative \(f'(x)\) tells you the immediate slope or rate of change.
• The second derivative \(f''(x)\) shows the rate at which the slope itself is changing.
• Higher derivatives like the third, fourth, and fifth provide even more detail, acting similar to how acceleration provides more context beyond mere speed.

In the Taylor polynomial, these higher-order derivatives evaluated at a given point help form each term in the series, refining the approximation and making it more accurate for values around \(x=9\).