Chapter 3, Problem 1RE

(a) Find $dy/dx$ by differentiating implicitly, (b) Solve the equation for $y$ as a function of $x$ , and find $dy/dx$ from that equation. (c) Confirm that the two results are consistent by expressing the derivative in part (a) as a function of $x$ alone.

$x^3+xy-2x=1$

To determine

To calculate: The derivative, $\frac{dy}{dx}$ , of the equation, $x^3+xy-2x=1$ , by differentiating implicitly.

Answer to Problem 1RE

The derivative, $\frac{dy}{dx}$ , of the equation, $x^3+xy-2x=1$ , is $\frac{2-y-3x^2}{x}$ .

Explanation of Solution

Calculation:

Differentiate the equation, $x^3+xy-2x=1$ , with respect to $x$ .

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

Hence, the derivative, $\frac{dy}{dx}$ , of the equation, $x^3+xy-2x=1$ , is $\frac{2-y-3x^2}{x}$ .

To determine

To calculate: The equation, $x^3+xy-2x=1$ , for $y$ as a function of $x$ , and then find $\frac{dy}{dx}$ .

Answer to Problem 1RE

The derivative, $\frac{dy}{dx}$ , is $-\frac{1}{x^2}-2x$ .

Explanation of Solution

Calculation:

Consider the provided equation, $x^3+xy-2x=1$ .

Simplify the equation for $y$ as a function of $x$ .

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

Differentiate $y$ with respect to $x$ .

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

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

To determine

To calculate: That the two results are consistent by expressing the derivative in part (a) as a function of $x$ alone.

Answer to Problem 1RE

The derivative, $\frac{dy}{dx}$ , is $-\frac{1}{x^2}-2x$ . So, this is the function of $x$ alone.

Explanation of Solution

Calculation:

From part (a),

$\frac{dy}{dx}=\frac{\left(2-y-3x^2\right)}{x}$ …… (1)

Simplify the equation, $x^3+xy-2x=1$ .

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

Substitute $y=\frac{1}{x}+2-x^2$ in equation (1).

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

Hence, the derivative, $\frac{dy}{dx}$ , is $-\frac{1}{x^2}-2x$ . So, this is the function of $x$ alone.