Question

if x = y (1+logx) find dy/dx​

Answer 1

Step-by-step explanation:

We have,

$x = y(1 + log(x) )$

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

$1 = (1 + log(x) ) \frac{dy}{dx} + y(1 + \frac{1}{x} )$

$\implies1 - \frac{y(x + 1)}{x} = (1 + log(x) ) \frac{dy}{dx}$

$\implies1 - \frac{x + 1}{1 + log(x) } = (1 + log(x) ) \frac{dy}{dx}$

$\implies \frac{dy}{dx} = \frac{1 + log(x) - x - 1}{(1 + log(x) )^{2} }$

$\implies \frac{dy}{dx} = \frac{ log(x) - x}{(1 + log(x)) ^{2} }$