Question

if x^y -y^x =0 , then dy/dx is ​

Answer 1

Answer ::

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$getting \: log \: both \: sides$

$y = {x}^{y} - {y}^{x} = 0$

$ylog = log {x}^{y} - log {y}^{x}$

$ylog = ylogx - xlogy$

$differentiating \: both \:sides$

$\frac{1}{y} \frac{dy}{dx} = y \times \frac{1}{x} \times 1 - logx \frac{dy}{dx} - (x \times \frac{l}{y} \times \frac{dy}{dx} - 1 \times logy)$

$\frac{dy}{dx} = y \bigg( \frac{y}{x} - logx \frac{dy}{dx} - ( \frac{x}{y} \frac{dy}{dx} - logy) \bigg)$

$put \: the \: value \: of \: y \: in \: equation \: as \: we \: know \: that \: this \: is \: implict \: function$

$\frac{dy}{dx} = {x}^{y} - {y}^{x} \bigg( \frac{y}{x} - logx \frac{dy}{dx} - \frac{x}{y} + logy \bigg) \: is \: the \: correct \: answer$