Find frac{dy}{dx},if y^{x}+x^{y}+x^{x}=a^{b}. | vedclass

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(A) Given that $y^{x}+x^{y}+x^{x}=a^{b}$.Let $u=y^{x}, v=x^{y}$ and $w=x^{x}$,then $u+v+w=a^{b}$.Differentiating with respect to $x$,we get $\frac{du}

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$\frac{-\left[y^{x} \log y+y \cdot x^{y-1}+x^{x}(1+\log x)\right]}{x \cdot y^{x-1}+x^{y} \log x}$

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(A) Given that $y^{x}+x^{y}+x^{x}=a^{b}$.Let $u=y^{x}, v=x^{y}$ and $w=x^{x}$,then $u+v+w=a^{b}$.Differentiating with respect to $x$,we get $\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}=0$ ... $(1)$.For $u=y^{x}$,taking log on both sides: $\log u = x \log y$.Differentiating: $\frac{1}{u} \frac{du}{dx} = x \cdot \frac{1}{y} \frac{dy}{dx} + \log y \cdot 1 \implies \frac{du}{dx} = y^{x} \left( \frac{x}{y} \frac{dy}{dx} + \log y \right)$ ... $(2)$.For $v=x^{y}$,taking log on both sides: $\log v = y \log x$.Differentiating: $\frac{1}{v} \frac{dv}{dx} = y \cdot \frac{1}{x} + \log x \cdot \frac{dy}{dx} \implies \frac{dv}{dx} = x^{y} \left( \frac{y}{x} + \log x \frac{dy}{dx} \right)$ ... $(3)$.For $w=x^{x}$,taking log on both sides: $\log w = x \log x$.Differentiating: $\frac{1}{w} \frac{dw}{dx} = x \cdot \frac{1}{x} + \log x \cdot 1 \implies \frac{dw}{dx} = x^{x}(1 + \log x)$ ... $(4)$.Substituting $(2), (3), (4)$ into $(1)$:$y^{x} \left( \frac{x}{y} \frac{dy}{dx} + \log y \right) + x^{y} \left( \frac{y}{x} + \log x \frac{dy}{dx} \right) + x^{x}(1 + \log x) = 0$.Grouping $\frac{dy}{dx}$ terms: $\frac{dy}{dx} (x \cdot y^{x-1} + x^{y} \log x) = -[y^{x} \log y + y \cdot x^{y-1} + x^{x}(1 + \log x)]$.Thus,$\frac{dy}{dx} = \frac{-[y^{x} \log y + y \cdot x^{y-1} + x^{x}(1 + \log x)]}{x \cdot y^{x-1} + x^{y} \log x}$.

Find $\frac{dy}{dx}$,if $y^{x}+x^{y}+x^{x}=a^{b}$.

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