If y^x + x^y + x^x = a^b, find dy dx

Question

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

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Answer

Given that $y^x + x^y + x^x = a^b$
Putting $u = y^x, v = x^y$ and $w = x^x$, we get $u + v + w = a^b$
Therefore $\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}+\frac{\text{dw}}{\text{dx}}=0\ .....(\text{i})$
Now, $u = y^x$. Taking logrithm on both sides, we have
$\log\text{u}=\text{x}\log\text{y}$
Differentiating both sides w.r.t. x, we have
$\frac{1}{\text{u}}\times\frac{\text{du}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$
$=\text{x}\frac{1}{\text{y}}\times\frac{\text{dy}}{\text{dx}}+\log\text{y}\times1$
So, $\frac{\text{du}}{\text{dx}}=\text{u}\Big(\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big)=\text{y}^\text{x}\Big[\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big]\ .....(\text{ii})$
Also $v = x^y$
Taking logarithm on both sides, we have
$\log\text{v}=\text{y}\log\text{x}$
Differentiating both sides w.r.t. x, we have
$\frac{1}{\text{v}}\times\frac{\text{dv}}{\text{dx}}=\text{y}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{dy}}{\text{dx}}$
$=\text{y}\times\frac{1}{\text{x}}+\log\text{x}\times\frac{\text{dy}}{\text{dx}}$
So, $\frac{\text{dv}}{\text{dx}}=\text{v}\Big[\frac{\text{y}}{\text{x}}+\log\text{x}\times\frac{\text{dy}}{\text{dx}}\Big]$
$=\text{x}^\text{y}\Big[\frac{\text{y}}{\text{x}}+\log\text{x}\times\frac{\text{dy}}{\text{dx}}\Big]$
Again $w = x^x$
Taking logarithm on both sides, we have
$\log\text{w}=\text{x}\log\text{x}$
Differentiating both sides w.r.t x, we have
$\frac{1}{\text{w}}\times\frac{\text{dw}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{x}}{\text{dx}}(\text{x})$
$=\text{x}.\frac{1}{\text{x}}+\log\text{x}\times1$
i.e. $\frac{\text{dw}}{\text{dx}}=\text{w}(1+\log\text{x})$
$=\text{x}^\text{x}(1+\log\text{x})\ .....(\text{iv})$
From (i), (ii), (iii), (iv), we have
$\text{y}^\text{x}\Big(\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big)+\text{x}^\text{y}\Big(\frac{\text{y}}{\text{x}}+\log\text{x}\frac{\text{dy}}{\text{dx}}\Big) \\ +\text{x}^\text{x}(1+\log\text{x})=0$
$\big(\text{x}\times\text{y}^{\text{x}-1}+\text{x}^\text{y}\times\log\text{x}\big) \\ \frac{\text{dy}}{\text{dx}}=-\text{x}^\text{x}(1+\log\text{x})-\text{y}\times\text{x}^{\text{y}-1}-\text{y}^\text{x}\log\text{y}=0$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{-\big[\text{y}^\text{x}\log\text{y}+\text{y}\times\text{x}^{\text{y}-1}+\text{x}^\text{x}(1+\log\text{x})\big]}{\text{x}\times\text{y}^{\text{x}-1}+\text{x}^\text{y}\log\text{y}}$