Chapter 3.2, Problem 7E

To determine

To find: The differentiation of the function $g\left(x\right)=\frac{1+2x}{3-4x}$.

Answer to Problem 7E

The differentiation of the function $g\left(x\right)=\frac{1+2x}{3-4x}$ is $g^{\prime }\left(x\right)=\frac{10}{{\left(3-4x\right)}^2}$.

Explanation of Solution

Given:

The function $g\left(x\right)=\frac{1+2x}{3-4x}$.

Derivative rule:

(1) Quotient Rule: If $f_1\left(x\right)$ and $f_2\left(x\right)$ are both differentiable, then

$\frac{d}{dx}\left[\frac{f_1\left(x\right)}{f_2\left(x\right)}\right]=\frac{f_2\left(x\right)\frac{d}{dx}\left[f_1\left(x\right)\right]-f_1\left(x\right)\frac{d}{dx}\left[f_2\left(x\right)\right]}{{\left[f_2\left(x\right)\right]}^2}$.

(2) Power Rule: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

(3) Sum rule: $\frac{d}{dx}\left(f+g\right)=\frac{d}{dx}\left(f\right)+\frac{d}{dx}\left(g\right)$

(4) Constant multiple rule: $\frac{d}{dx}\left(cf\right)=c\frac{d}{dx}\left(f\right)$

(5) Difference rule: $\frac{d}{dx}\left(f-g\right)=\frac{d}{dx}\left(f\right)-\frac{d}{dx}\left(g\right)$

Calculation:

The derivative of the function is $g^{\prime }\left(x\right)$, which is obtained as follows.

$\begin{array}{c}{g}^{\prime }\left(x\right)=\frac{d}{dx}\left(g\left(x\right)\right)\\ =\frac{d}{dx}\left(\frac{1+2x}{3-4x}\right)\end{array}$

Substitute $1+2x$ for $f_1\left(x\right)$ and $3-4x$ for $f_2\left(x\right)$ in the quotient rule (1),

$g^{\prime }\left(x\right)=\frac{\left(3-4x\right)\frac{d}{dx}\left[1+2x\right]-\left(1+2x\right)\frac{d}{dx}\left[3-4x\right]}{{\left[3-4x\right]}^2}$

Apply the derivative rules (3), (4), and (5),

$\begin{array}{c}{g}^{\prime }\left(x\right)=\frac{\left(3-4x\right)\left[\frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(2x\right)\right]-\left(1+2x\right)\left[\frac{d}{dx}\left(3\right)-\frac{d}{dx}\left(4x\right)\right]}{{\left(3-4x\right)}^2}\\ =\frac{\left(3-4x\right)\left[\frac{d}{dx}\left(1\right)+2\frac{d}{dx}\left(x\right)\right]-\left(1+2x\right)\left[\frac{d}{dx}\left(3\right)-4\frac{d}{dx}\left(x\right)\right]}{{\left(3-4x\right)}^2}\end{array}$

Apply the power rule (2) and simplify the terms,

$\begin{array}{c}{g}^{\prime }\left(x\right)=\frac{\left(3-4x\right)\left[0+2\left(1\right)\right]-\left(1+2x\right)\left[0-4\left(1\right)\right]}{{\left(3-4x\right)}^2}\\ =\frac{\left(3-4x\right)\left[2\right]-\left(1+2x\right)\left[-4\right]}{{\left(3-4x\right)}^2}\\ =\frac{6-8x-\left(-4-8x\right)}{{\left(3-4x\right)}^2}\\ =\frac{10}{{\left(3-4x\right)}^2}\end{array}$

Therefore, the differentiation of the function $g\left(x\right)=\frac{1+2x}{3-4x}$ is $g^{\prime }\left(x\right)=\frac{10}{{\left(3-4x\right)}^2}$.