If {y^x} + {x^y} = {a^b},then frac{dy}{dx} = | vedclass

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(A) Given the equation: ${y^x} + {x^y} = {a^b}$.Let $u = {x^y}$ and $v = {y^x}$.Then $u + v = {a^b}$.Differentiating with respect to $x$,we get $\frac

Vedclass · Accepted Answer

$ - \frac{y{x^{y - 1}} + {y^x}\log y}{x{y^{x - 1}} + {x^y}\log x}$

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(A) Given the equation: ${y^x} + {x^y} = {a^b}$.Let $u = {x^y}$ and $v = {y^x}$.Then $u + v = {a^b}$.Differentiating with respect to $x$,we get $\frac{du}{dx} + \frac{dv}{dx} = 0$.For $u = {x^y}$,taking log on both sides: $\log u = y \log x$.Differentiating: $\frac{1}{u} \frac{du}{dx} = y \cdot \frac{1}{x} + \log x \cdot \frac{dy}{dx} \implies \frac{du}{dx} = {x^y} \left( \frac{y}{x} + \log x \frac{dy}{dx} \right) = y{x^{y - 1}} + {x^y} \log x \frac{dy}{dx}$.For $v = {y^x}$,taking log on both sides: $\log v = x \log y$.Differentiating: $\frac{1}{v} \frac{dv}{dx} = x \cdot \frac{1}{y} \frac{dy}{dx} + \log y \cdot 1 \implies \frac{dv}{dx} = {y^x} \left( \frac{x}{y} \frac{dy}{dx} + \log y \right) = x{y^{x - 1}} \frac{dy}{dx} + {y^x} \log y$.Substituting these into $\frac{du}{dx} + \frac{dv}{dx} = 0$:$y{x^{y - 1}} + {x^y} \log x \frac{dy}{dx} + x{y^{x - 1}} \frac{dy}{dx} + {y^x} \log y = 0$.$\frac{dy}{dx} ({x^y} \log x + x{y^{x - 1}}) = -(y{x^{y - 1}} + {y^x} \log y)$.$\frac{dy}{dx} = - \frac{y{x^{y - 1}} + {y^x} \log y}{x{y^{x - 1}} + {x^y} \log x}$.

If ${y^x} + {x^y} = {a^b}$,then $\frac{dy}{dx} = $

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