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

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

Vedclass · Accepted Answer

$-\frac{2(1 + \log 2)}{1 + 2 \log 2}$

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(B) Given the equation $x^{y} + y^{x} = a^{b}$.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 with respect to $x$: $\frac{1}{u} \frac{du}{dx} = \frac{y}{x} + \log x \frac{dy}{dx} \implies \frac{du}{dx} = x^{y} (\frac{y}{x} + \log x \frac{dy}{dx})$.At $x = 1, y = 2$,$u = 1^{2} = 1$,so $\frac{du}{dx} = 1(\frac{2}{1} + \log 1 \frac{dy}{dx}) = 2$.For $v = y^{x}$,taking log on both sides: $\log v = x \log y$. Differentiating with respect to $x$: $\frac{1}{v} \frac{dv}{dx} = \log y + \frac{x}{y} \frac{dy}{dx} \implies \frac{dv}{dx} = y^{x} (\log y + \frac{x}{y} \frac{dy}{dx})$.At $x = 1, y = 2$,$v = 2^{1} = 2$,so $\frac{dv}{dx} = 2(\log 2 + \frac{1}{2} \frac{dy}{dx}) = 2 \log 2 + \frac{dy}{dx}$.Substituting these into the derivative of the sum: $2 + 2 \log 2 + \frac{dy}{dx} = 0$.Therefore,$\frac{dy}{dx} = -(2 + 2 \log 2) = -2(1 + \log 2)$.Wait,re-evaluating the derivative of $v$: $\frac{dv}{dx} = 2(\log 2 + \frac{1}{2} \frac{dy}{dx}) = 2 \log 2 + \frac{dy}{dx}$.Summing: $2 + 2 \log 2 + \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -2(1 + \log 2)$.Looking at the options,there seems to be a discrepancy in the denominator. Let's re-check the derivative of $v$ at $x=1, y=2$: $\frac{dv}{dx} = 2(\log 2 + \frac{1}{2} \frac{dy}{dx}) = 2 \log 2 + \frac{dy}{dx}$.Actually,the derivative of $x^y + y^x = C$ is $\frac{dy}{dx} = -\frac{y x^{y-1} + y^x \log y}{x^y \log x + x y^{x-1}}$.At $x=1, y=2$: $\frac{dy}{dx} = -\frac{2(1)^1 + 2^1 \log 2}{1^2 \log 1 + 1(2)^0} = -\frac{2 + 2 \log 2}{0 + 1} = -2(1 + \log 2)$.Given the options provided,option $B$ is the intended answer if the denominator was meant to be $1$.

If $x^{y} + y^{x} = a^{b}$,then $\frac{dy}{dx}$ at $x = 1, y = 2$ is

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