91Ó°ÊÓ

In calculus, derivatives help us understand how a function changes as its input changes. This concept is essential for developing Taylor polynomials, which represent a function as a series of derivatives evaluated at a certain point. Let's break down the derivatives used in our exercise.

When computing the derivatives of the function \(f(x) = (x+5)^{1/3}\), we are examining how the function behaves at points close to \(x = 3\). Differentiate step-by-step:

• **First Derivative (\(f'(x)\))**: We use the power rule and chain rule to find the first derivative \(f'(x) = \frac{1}{3}(x+5)^{-2/3}\), and then evaluate it at \(x = 3\), yielding \(f'(3) = \frac{1}{12}\).
• **Second Derivative (\(f''(x)\))**: Differentiate \(f'(x)\) to get \(f''(x) = \frac{-2}{9}(x+5)^{-5/3}\) and evaluate at \(x = 3\), resulting in \(f''(3) = \frac{-1}{144}\).
• **Third Derivative (\(f'''(x)\))**: Further differentiate \(f''(x)\) to find \(f'''(x) = \frac{10}{27}(x+5)^{-8/3}\) and evaluate it at \(x = 3\), which gives \(f'''(3) = \frac{5}{3456}\).

These derivatives are crucial in constructing the Taylor polynomial, as they provide the rates of change needed to approximate our function around the point \(c = 3\).

Function evaluation involves computing the output of a function for a specific input. In the context of Taylor Polynomials, evaluating the function at the base point \(c\) is a fundamental step.

In this exercise, we evaluate the function \(f(x) = (x+5)^{1/3}\) at \(c = 3\). This is the first term in our Taylor polynomial and represents the value of the function at this point.
For our specific case:

• Calculate \(f(3)\) by plugging in \(3\) into the function: \(f(3) = (3+5)^{1/3} = 8^{1/3} = 2\).

This calculation provides the initial value for the Taylor polynomial \(T_3(x)\), ensuring that our polynomial matches the function at the given base point. This correspondence is the foundation upon which the entire polynomial is built, capturing the function's value, slope, and higher-order changes at \(c\).

The chain rule is a fundamental tool in calculus used for differentiating composite functions. It allows us to break down complex derivatives into simpler parts, making it easier to tackle functions like \(f(x) = (x+5)^{1/3}\).

In this example, we must differentiate a composite function, which is the composition of two functions: the inner function \(x+5\) and the outer function \((x+5)^{1/3}\). The chain rule provides a structured way to find derivatives:

• For \(f(x) = (x+5)^{1/3}\), we note that the outer function is raising to the power \(1/3\).
• The inner function is \(x+5\).
• The derivative using the chain rule is:\[f'(x) = \frac{1}{3}(x+5)^{-2/3} \cdot \frac{d}{dx}(x+5) = \frac{1}{3}(x+5)^{-2/3} \cdot 1 = \frac{1}{3}(x+5)^{-2/3}\]

By applying the chain rule, we can efficiently find the derivatives necessary for forming the Taylor polynomial. This allows us to approximate complex functions with simple polynomial expressions that are easier to analyze and compute.