Chapter 3.3, Problem 14E

(a)

To determine

To calculate: The value of $\frac{dy}{dx}$ by applying the Quotient Rule.

Answer to Problem 14E

The required derivative is $1-\frac{3}{x^2}$ .

Explanation of Solution

Given information:

The curve: $y=\frac{\left(x^3+3\right)}{x}$

Formula used:

Quotient rule:

  $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u^{\prime }v-v^{\prime }u}{v^2}$

Calculation:

The given curve is $y=\frac{\left(x^3+3\right)}{x}$ .

Differentiate the given curve using the Quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u^{\prime }v-v^{\prime }u}{v^2}$ .

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left[\frac{x^2+3}{x}\right]\\ \frac{dy}{dx}=\frac{x\frac{d}{dx}\left(x^2+3\right)-\left(x^2+3\right)\frac{d}{dx}\left(x\right)}{x^2}\\ \frac{dy}{dx}=\frac{x\left(2x+0\right)-\left(x^2+3\right)\left(1\right)}{x^2}\\ \frac{dy}{dx}=\frac{2x^2-x^2-3}{x^2}\end{array}$

Simplify the above expression further.

  $\begin{array}{c}\frac{dy}{dx}=\frac{2x^2-x^2-3}{x^2}\\ \frac{dy}{dx}=\frac{x^2-3}{x^2}\\ \frac{dy}{dx}=\frac{x^2}{x^2}-\frac{3}{x^2}\\ \frac{dy}{dx}=1-\frac{3}{x^2}\end{array}$

Hence, the required derivative is $1-\frac{3}{x^2}$ .

(b)

To determine

To calculate: The value of $\frac{dy}{dx}$ by dividing the terms in the numerator by the denominator.

Answer to Problem 14E

The required derivative is $1-\frac{3}{x^2}$ .

Explanation of Solution

Given information:

The curve: $y=\frac{\left(x^3+3\right)}{x}$

Calculation:

The given curve is $y=\frac{\left(x^3+3\right)}{x}$ .

Rewrite the given curve as:

  $\begin{array}{l}y=\frac{x^2+3}{x}\\ y=\frac{x^2}{x}+\frac{3}{x}\\ y=x+3x^{-1}\end{array}$

Differentiate the above expression.

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

Hence, the required derivative is $1-\frac{3}{x^2}$ .