Question

if 4x+ 3y= log 4x- 3y find the dy÷ dx​

Answer 1

Answer:

$4x + 3y = log(4x - 3y)$

Differentiating w.r.t. x, we get,

$4 + 3 \times \frac{dy}{dx} = \frac{1}{4x - 3y} (4 - 3\times \frac{dy}{dx} )$

$4 + 3 \times \frac{dy}{dx} = \frac{4}{4x - 3y} - \frac{3}{4x - 3y} \frac{dy}{dx}$

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

$\frac{dy}{dx} = \frac{ \frac{4}{4x - 3y} - 4 }{3 + \frac{3}{4x - 3y} }$

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