Chapter 3.7, Problem 21E

To determine

To calculate: The derivative $\left(y\text{'}\right)$ and double derivative $\left(y\text{'}\text{'}\right)$ of the function $y=x^2\ln \left(2x\right)$ .

Answer to Problem 21E

The derivative $\left(y\text{'}\right)$ of the function is $x+2x\ln \left(2x\right)$ and double derivative $\left(y\text{'}\text{'}\right)$ is $2\ln \left(2x\right)+3$ .

Explanation of Solution

Given information:

The function $y=x^2\ln \left(2x\right)$ .

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 $y=x^2\ln \left(2x\right)$ .

Differentiate both sides with respect to x ,

  $\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(x^2\ln \left(2x\right)\right)$

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.

Apply it,

  $\begin{array}{c}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(x^2\ln \left(2x\right)\right)\\ y\text{'}=x^2\cdot \frac{1}{2x}\cdot \left(2\right)+2x\cdot \ln \left(2x\right)\\ y\text{'}=x+2x\ln \left(2x\right)\end{array}$

To find double derivative, again differentiate the function with respect to x ,

  $\begin{array}{c}y\text{'}=x+2x\ln \left(2x\right)\\ \frac{d}{dx}\left(y\text{'}\right)=\frac{d}{dx}\left(x+2x\ln \left(2x\right)\right)\end{array}$

Recall the 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.

Apply it,

  $\begin{array}{c}\frac{d}{dx}\left(y\text{'}\right)=\frac{d}{dx}\left(x+2x\ln \left(2x\right)\right)\\ y\text{'}\text{'}=1+2x\frac{1}{2x}\left(2\right)+2\ln \left(2x\right)\\ y\text{'}\text{'}=1+2+2\ln \left(2x\right)\\ y\text{'}\text{'}=2\ln \left(2x\right)+3\end{array}$

Thus, the derivative $\left(y\text{'}\right)$ of the function is $x+2x\ln \left(2x\right)$ and double derivative $\left(y\text{'}\text{'}\right)$ is $2\ln \left(2x\right)+3$ .