Chapter 3.7, Problem 46E

To determine

To calculate: The $9\text{th}$ derivative of the function $x^8\ln x$ .

Answer to Problem 46E

The $9\text{th}$ derivative of the function $x^8\ln x$ is $\frac{d^9}{d{x}^9}\left(x^8\ln x\right)=\frac{8!}{x}$ .

Explanation of Solution

Given information:

The function $x^8\ln x$ .

Formula used:

Thechain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

Power rule for differentiation is $\frac{d}{dx}{x}^n=n{x}^{n-1}$ .

Product rule for differentiation is $\frac{d}{dx}\left(f\cdot g\right)=f\text{'}\left(x\right)g\left(x\right)+f\left(x\right)g\text{'}\left(x\right)$ where f and g are functions of x .

Calculation:

Consider the function $f\left(x\right)=x^8\ln x$ .

Differentiate both sides with respect to x ,

  $\frac{d}{dx}\left(f\left(x\right)\right)=\frac{d}{dx}\left(x^8\ln x\right)$

Recall that power rule for differentiation is $\frac{d}{dx}{x}^n=n{x}^{n-1}$ and chain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

Also for the terms of the above expression, apply the product rule for differentiation.

Recall that product rule for differentiation is $\frac{d}{dx}\left(f\cdot g\right)=f\text{'}\left(x\right)g\left(x\right)+f\left(x\right)g\text{'}\left(x\right)$ where f and g are functions of x .

Apply it.

  $\begin{array}{c}f\text{'}\left(x\right)=x^8\left(\ln x\right)\text{'}+\left(x^8\right)\text{'}\ln x\\ f\text{'}\left(x\right)=8x^7\ln x+x^8\frac{1}{x}\\ f\text{'}\left(x\right)=8x^7\ln x+x^7\end{array}$

Similarly, for the second derivative,apply the product rule for differentiation that is $\frac{d}{dx}\left(f\cdot g\right)=f\text{'}\left(x\right)g\left(x\right)+f\left(x\right)g\text{'}\left(x\right)$ where f and g are functions of x .

  $\begin{array}{c}f\text{'}\text{'}\left(x\right)=8\cdot 7x^6\ln x+8x^7\cdot \frac{1}{x}+7x^6\\ f\text{'}\text{'}\left(x\right)=8\cdot 7x^6\ln x+8x^6+7x^6\end{array}$

Similarly, find the third derivative,apply the product rule for differentiation that is $\frac{d}{dx}\left(f\cdot g\right)=f\text{'}\left(x\right)g\left(x\right)+f\left(x\right)g\text{'}\left(x\right)$ where f and g are functions of x .

  $\begin{array}{c}f\text{'}\text{'}\text{'}\left(x\right)=8\cdot 7\cdot 6x^5\ln x+8\cdot 7x^6\cdot \frac{1}{x}+8\cdot 6x^5+7\cdot 6x^5\\ f\text{'}\text{'}\left(x\right)=8\cdot 7\cdot 6x^5\ln x+8\cdot 7x^5+48x^5+42x^5\end{array}$

When derivatives are evaluated in same pattern it is observed that all terms except the first term will be a constant number in the eight derivative.

It is known that derivative of a constant term is zero. So, in the ninth derivative, all terms except the first term becomes zero.

Now, eighth derivative of the function is,

  $f^8\left(x\right)=8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1x^0\ln x+\text{constant}$

And ninth derivative is,

  $\begin{array}{c}{f}^9\left(x\right)=8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\frac{1}{x}\\ =\frac{8!}{x}\end{array}$

Thus, the $9\text{th}$ derivative of the function $x^8\ln x$ is $\frac{d^9}{d{x}^9}\left(x^8\ln x\right)=\frac{8!}{x}$ .