Find frac{dy}{dx} for the function x^{y} + y^ | vedclass

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(A) The given function is $x^{y} + y^{x} = 1$.Let $u = x^{y}$ and $v = y^{x}$.Then the equation becomes $u + v = 1$.Differentiating with respect to $x

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given function is $x^{y} + y^{x} = 1$.Let $u = x^{y}$ and $v = y^{x}$.Then the equation becomes $u + v = 1$.Differentiating with respect to $x$,we get $\frac{du}{dx} + \frac{dv}{dx} = 0$  $(1)$.For $u = x^{y}$,taking log on both sides: $\log u = y \log x$.Differentiating with respect to $x$: $\frac{1}{u} \frac{du}{dx} = y \cdot \frac{1}{x} + \log x \cdot \frac{dy}{dx}$.So,$\frac{du}{dx} = x^{y} \left( \frac{y}{x} + \log x \frac{dy}{dx} \right) = y x^{y-1} + x^{y} \log x \frac{dy}{dx}$  $(2)$.For $v = y^{x}$,taking log on both sides: $\log v = x \log y$.Differentiating with respect to $x$: $\frac{1}{v} \frac{dv}{dx} = x \cdot \frac{1}{y} \frac{dy}{dx} + \log y \cdot 1$.So,$\frac{dv}{dx} = y^{x} \left( \frac{x}{y} \frac{dy}{dx} + \log y \right) = x y^{x-1} \frac{dy}{dx} + y^{x} \log y$  $(3)$.Substituting $(2)$ and $(3)$ into $(1)$:$y x^{y-1} + x^{y} \log x \frac{dy}{dx} + x y^{x-1} \frac{dy}{dx} + y^{x} \log y = 0$.$\frac{dy}{dx} (x^{y} \log x + x y^{x-1}) = -(y x^{y-1} + y^{x} \log y)$.Therefore,$\frac{dy}{dx} = -\frac{y x^{y-1} + y^{x} \log y}{x^{y} \log x + x y^{x-1}}$.

Find $\frac{dy}{dx}$ for the function $x^{y} + y^{x} = 1$.

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