Question

If xʸ=eˣ-ʸ,prove dy/dx=logx/(logx+1)²

Answer 1

Step-by-step explanation:

${x}^{y} = {e}^{x - y}$

Taking log both sides

$y \: log \: x = x - y...eq1 \\ \\ y \: log \: x + y = x \\ \\ y(1 + log \: x) = x \\ \\ \frac{y}{x} = \frac{1}{1 + log \: x} ...eq2$

Differentiate both side with respect to x

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

From eq1

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

From eq2,put the value of y/x

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

Hence proved.

Hope it helps you.