Chapter 3.3, Problem 19E

To determine

To calculate: The value of $\frac{dy}{dx}$ .

Answer to Problem 19E

The required value of $\frac{dy}{dx}$ is $\frac{3}{x^4}$

Explanation of Solution

Given information:

The expression: $y=\frac{\left(x-1\right)\left(x^2+x+1\right)}{x^3}$

Formula used:

Power rule:

  $\frac{d}{dx}{x}^n=n{x}^{n-1}$

Calculation:

The given expression is $y=\frac{\left(x-1\right)\left(x^2+x+1\right)}{x^3}$ .

Rewrite the above expression as:

  $\begin{array}{c}y=\frac{\left(x-1\right)\left(x^2+x+1\right)}{x^3}\\ y=\frac{x\left(x^2+x+1\right)-1\left(x^2+x+1\right)}{x^3}\\ y=\frac{x^3+\cancel{x^2}+\cancel{x}-\cancel{x^2}-\cancel{x}-1}{x^3}\\ y=\frac{x^3-1}{x^3}\end{array}$

Simplify the expression further.

  $\begin{array}{c}y=\frac{x^3-1}{x^3}\\ y=\frac{\cancel{x^3}}{\cancel{x^3}}-\frac{1}{x^3}\\ y=1-\frac{1}{x^3}\\ y=1-x^{-3}\end{array}$

To find the value of $\frac{dy}{dx}$ , use the power rule on the above expression.

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(1-x^{-3}\right)\\ =\frac{d}{dx}\left(1\right)-\frac{d}{dx}\left(x^{-3}\right)\\ =0-\left(-3\right)x^{\left(-3-1\right)}\\ =3x^{-4}\end{array}$

Simplify the above value further.

  $\begin{array}{c}\frac{dy}{dx}=3x^{-4}\\ \frac{dy}{dx}=\frac{3}{x^4}\end{array}$

Hence, the required value of $\frac{dy}{dx}$ is $\frac{3}{x^4}$ .