Question
यदि ${y^x} + {x^y} = {a^b}$, तो $\frac{{dy}}{{dx}} = $
माना ${x^y} = u$ और ${y^x} = v$
==> $u + v = {a^b}$
==>$\frac{{du}}{{dx}} + \frac{{dv}}{{dx}} = 0$
दोनों का अलग-अलग $\log $ लेकर, अवकलन करने पर,
$\frac{{du}}{{dx}} = {x^y}\left( {\frac{y}{x} + \frac{{dy}}{{dx}}\log x} \right)$…..$(i)$
तथा $\frac{{dv}}{{dx}} = {y^x}\left( {\log y + \frac{x}{y}.\frac{{dy}}{{dx}}} \right)$…..$(ii)$
समी. $(i)$ व समी. $(ii)$ से,
$\frac{{dy}}{{dx}} = - \frac{{y{x^{y - 1}} + {y^x}\log y}}{{{x^y}\log x + x{y^{x - 1}}}}$.
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