Question

If log (x + y) = log(xy) + p, where p is a constant, then prove that ${\displaystyle \frac{\text{dy}}{\text{dx}}=\frac{-y^2}{x^2}}$.

Answer 1

log (x + y) = log(xy) + p
∴ log( x + y) = logx + logy + p
Differentiating both sides w.r.t. x, we get
${\displaystyle \frac{1}{x+y}.\frac{\text{d}}{\text{dx}}(x+y)=\frac{1}{x}+\frac{1}{y}.\frac{\text{dy}}{\text{dx}}+0}$

∴ ${\displaystyle \frac{1}{x+y}(1+\frac{\text{dy}}{\text{dx}})=\frac{1}{x}+\frac{1}{y}.\frac{\text{dy}}{\text{dx}}}$

∴ ${\displaystyle \frac{1}{x+y}+\frac{1}{x+y}.\frac{\text{dy}}{\text{dx}}=\frac{1}{x}+\frac{1}{y}.\frac{\text{dy}}{\text{dx}}}$

∴ ${\displaystyle (\frac{1}{x+y}-\frac{1}{y})\frac{\text{dy}}{\text{dx}}=\frac{1}{x}-\frac{1}{x+y}}$

∴ ${\displaystyle \left[\frac{y-x-y}{y(x+y)}\right]\frac{\text{dy}}{\text{dx}}=\frac{x+y-x}{x(x+y)}}$

∴ ${\displaystyle \left[\frac{-x}{y(x+y)}\right]\frac{\text{dy}}{\text{dx}}=\frac{y}{x(x+y)}}$

∴ ${\displaystyle (-\frac{x}{y})\frac{\text{dy}}{\text{dx}}=\frac{y}{x}}$

∴ ${\displaystyle \frac{\text{dy}}{\text{dx}}=-\frac{y^2}{x^2}}$.