Question

If 4x + 3y = log(4x – 3y), then find ${\displaystyle \frac{\text{dy}}{\text{dx}}}$

Answer 1

Given 4x + 3y = log(4x – 3y)

Differentiating both sides with respect to x,

${\displaystyle 4\left(1\right)+3\frac{\text{dy}}{\text{dx}}=\frac{1}{(4x-3y)}\frac{\text{d}}{\text{dx}}(4x-3y)}$

${\displaystyle 4+3\frac{\text{dy}}{\text{dx}}=\frac{1}{(4x-3y)}(4\left(1\right)-3\frac{\text{dy}}{\text{dx}})}$

${\displaystyle (4x-3y)(4+3\frac{\text{dy}}{\text{dx}})=4-3\frac{\text{dy}}{\text{dx}}}$

${\displaystyle 16x+12x\frac{\text{dy}}{\text{dx}}-12y-9y\frac{\text{dy}}{\text{dx}}=4-3\frac{\text{dy}}{\text{dx}}}$

${\displaystyle 12x\frac{\text{dy}}{\text{dx}}+3\frac{\text{dy}}{\text{dx}}-9y\frac{\text{dy}}{\text{dx}}=4-16x+12y}$

${\displaystyle \frac{\text{dy}}{\text{dx}}}$[12x + 3 - 9y] = 4[1 - 4x + 3y]

${\displaystyle \frac{\text{dy}}{\text{dx}}=\frac{4[1-4x+3y]}{3[4x+1-3y]}}$

= ${\displaystyle \frac{4[1-4x+3y]}{3[1+4x-3y]}}$