Question

Find y' if x^y = y^x​

Answer 1

The answer is $\boxed {\frac{dy}{dx} = \frac{y^{x}logy-yx^{y-1}}{x^{y}logx-xy^{x-1}}}$.

Apply logarithmic differentiation on each side.

LHS

• u = x^y
• log u = y log x
• 1/u du/dx = y/x + log x (dy/dx)
• du/dx = x^y (y/x + log x dy/dx)
• du/dx = yx^(y - 1) + x^y logx dy/dx

RHS

• v = y^x
• log v = x log y
• 1/v dv/dx = x/y dy/dx + logy
• dv/dx = y^x (x/y dy/dx + logy)
• dv/dx = xy^(x - 1) dy/dx + y^x logy

Equating both sides

• yx^(y - 1) + x^y logx dy/dx = xy^(x - 1) dy/dx + y^x logy
• dy/dx (x^y logx - xy^(x - 1)) = y^x logy - yx^(y - 1)
• $\boxed {\frac{dy}{dx} = \frac{y^{x}logy-yx^{y-1}}{x^{y}logx-xy^{x-1}}}$