Question

Find $\frac{dy}{dx}$ in each of the following
(i) $xy=c^2$
(ii) $4x+3y=\log (4x-3y)$

Answer 1

(i) $xy=c^2$

$y=\frac{c^2}{x}$

$\frac{dy}{dx}=-\frac{c^2}{x^2}$

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

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

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

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

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

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

$\Rightarrow \frac{dy}{dx}=\frac{4}{3}\left[\frac{4x-3y-1}{-4x+3y-1}\right]$