Find dy dx if y^x + x^y + x^x = a^b - Vidyadip

Question

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

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Answer

Let $u = {y^x},v = {x^y},w = {x^x}$, then$u + v + w = {a^b}$
Therefore, $\frac{{du}}{{dx}} + \frac{{dw}}{{dx}} + \frac{{dv}}{{dx}} = 0 ...(1)$
$u = {y^x}$
Taking $\log$ both sides, we get
$\log u = \log {y^x}$
$\log u = x.\log y$
Differentiate both sides $w.r.t.$ to $x$
$\frac{1}{u}.\frac{{du}}{{dx}} = x.\frac{1}{y}.\frac{{dy}}{{dx}} + \log y.1$
$\frac{{du}}{{dx}} = u\left[ {\frac{x}{y}.\frac{{dy}}{{dx}} + \log y} \right]$
$\frac{{du}}{{dx}} = {y^x}\left( {\frac{x}{y}.\frac{{dy}}{{dx}} + \log y} \right)...(2)$
$v = {x^y}$
Taking $\log$ both sides, we get
$\log v = \log {x^y}$
$\log v = y.\log x$
$\frac{1}{v}.\frac{{dv}}{{dx}} = y.\frac{1}{x} + \log x.\frac{{dy}}{{dx}}$
$\frac{{dv}}{{dx}} = v\left[ {\frac{y}{x} + \log x.\frac{{dy}}{{dx}}} \right]$
$\frac{{dv}}{{dx}} = {x^y}\left[ {\frac{y}{x} + \log x.\frac{{dy}}{{dx}}} \right]...(3)$
$w = {x^x}$
Taking $\log$ both sides, we get
$\log w = \log {x^x}$
$\log w = x\log x$
$\frac{1}{w}.\frac{{dw}}{{dx}} = x.\frac{1}{x} + \log x.1$
$\frac{1}{w}.\frac{{dw}}{{dx}} = 1 + \log x$
$\frac{{dw}}{{dx}} = w\left( {1 + \log x} \right)$
$\frac{{dw}}{{dx}} = {x^x}\left( {1 + \log x} \right)... (4)$
$\frac{{dy}}{{dx}} = \frac{{ - {x^x}\left( {1 + \log x} \right) - y.{x^{y - 1}} - {y^x}\log y}}{{x.{y^{x - 1}} + {x^y}\log x}}$ $($by putting $(2), (3)$ and $(4)$ in $(1))$

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