Problem 72 - Essential Calculus Early Transcendentals

Question

Find \(\frac{d^{3}}{d x^{9}}\left(x^{8} \ln x\right)\)

Step-by-Step Solution

Answer

The 9th derivative of \(x^8 \ln x\) is 0.

Step 1: Identify the Function and Derivation Order

We are tasked with finding the derivative of the function \(x^8 \ln x\) with respect to \(x\) for a total of 9 orders. This means we need to find \(\frac{d^{9}}{d x^{9}}(x^8 \ln x)\).

Step 2: Compute Initial Derivative (First Order)

Use the product rule to find the first derivative of \(x^8 \ln x\). Recall that the product rule is \((uv)' = u'v + uv'\). Set \(u = x^8\) and \(v = \ln x\). Then, \(u' = 8x^7\) and \(v' = \frac{1}{x}\).Calculate: \(\frac{d}{dx}(x^8 \ln x) = 8x^7 \ln x + x^8 \cdot \frac{1}{x} = 8x^7 \ln x + x^7.\)

Step 3: Simplify the First Derivative

The first derivative \(\frac{d}{dx}(x^8 \ln x)\) simplifies to \(8x^7 \ln x + x^7\).

Step 4: Evaluate Higher-Order Derivatives

We notice that the expression involves repeated derivatives of \(x^n \ln x\) and \(x^n\). In subsequent steps, observe that the derivative of \(x^k\) with respect to \(x\) is \(kx^{k-1}\) and repeats decrements of powers with each derivative.For each order, the logarithmic term vanishes after deriving 8 times, at the ninth derivative as no cross-term will remain with a positive power of \(x\).

Step 5: Conclude Evaluation at Ninth Derivative

By the ninth derivative, both terms, \(8x^7 \ln x\) and \(x^7\), will have all vanished (i.e., become zero) since deriving beyond the nth power of a polynomial term eventually yields zero.Therefore, \(\frac{d^{9}}{d x^{9}}(x^8 \ln x) = 0\).

Key Concepts

Product Rule

The product rule is an essential tool in calculus for finding the derivative of a function that is a product of two other functions.
If you have a function of the form \( f(x) = u(x)v(x) \), the product rule states that the derivative \( f'(x) \) is given by:

• \((uv)' = u'v + uv'\).

For example, in the original exercise, you have the function \(x^8 \ln x\). You can think of this as \(u = x^8\) and \(v = \ln x\).
By applying the product rule:

• \(u' = 8x^7\)
• \(v' = \frac{1}{x}\)
• The derivative \(f'(x) = 8x^7 \ln x + x^8 \cdot \frac{1}{x} = 8x^7 \ln x + x^7\)

Using the product rule correctly simplifies the effort needed to find derivatives of products of functions, particularly when you have repeated derivatives, as in this exercise.

Polynomial Derivatives

Understanding derivatives of polynomials is a fundamental aspect of calculus and greatly facilitates handling more complex differential problems.
The derivative of a simple polynomial \(x^n\) with respect to \(x\) is given by the power rule:

• \(\frac{d}{dx}(x^n) = nx^{n-1}\).

This rule is applied repeatedly when calculating higher-order derivatives, which are derivatives taken multiple times.
For instance, for any term \(x^n\), each time you differentiate it, the power \(n\) decreases by one.Consistent application, such as in the given exercise, simplifies to zero when derived beyond the power it initially holds.
Calculating derivatives consecutively, like the 9th derivative of a term \(x^8\), results in zero because the repeated application of the power rule quickly reduces the term \(x^n\) to a constant and eventually to zero.

Natural Logarithm Differentiation

Natural logarithm functions commonly appear in calculus problems and require specialized rules for differentiation.
The standard differentiation rule for the natural logarithm \(\ln x\) is:

• \(\frac{d}{dx} (\ln x) = \frac{1}{x}\).

In combinations like \(u(x) \ln x\), as seen in the provided exercise, this becomes part of a product differentiation rule usage.
When differentiating \(x^8 \ln x\), \(\ln x\) is treated as one part of the product.The method involves differentiating each part separately and then combining them according to the product rule.
While this example specifically dealt with a product involving \(\ln x\), the technique applies to more complex logarithmic differentiations, enabling handling of tougher calculus challenges.