Let ${x^y} = u$ and ${y^x} = v$
==> $u + v = {a^b}$
==>$\frac{{du}}{{dx}} + \frac{{dv}}{{dx}} = 0$
Now differentiating both by taking their $\log $ separately
$\frac{{du}}{{dx}} = {x^y}\left( {\frac{y}{x} + \frac{{dy}}{{dx}}\log x} \right)$ …..$(i)$
and $\frac{{dv}}{{dx}} = {y^x}\left( {\log y + \frac{x}{y}.\frac{{dy}}{{dx}}} \right)$ …..$(ii)$
Therefore, by $(i)$ and $(ii),$
$\frac{{dy}}{{dx}} = - \frac{{y{x^{y - 1}} + {y^x}\log y}}{{{x^y}\log x + x{y^{x - 1}}}}$.
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$\overrightarrow{ r }=\hat{ i }+\lambda(-\hat{ i }+2 \hat{ j }+2 \hat{ k }), \lambda \in R \text { and }$
$\overrightarrow{ r }=\mu(2 \hat{ i }-\hat{ j }+2 \hat{ k }), \mu \in R$
respectively. If $L _3$ is a line which is perpendicular to both $L _1$ and $L _2$ and cuts both of them, then which of the following options describe(s) $L _3$ ?
$(1)$ $\overrightarrow{ r }=\frac{1}{3}(2 \hat{ i }+\hat{ k })+ t (2 \hat{ i }+2 \hat{ j }-\hat{ k }), t \in R$
$(2)$ $\overrightarrow{ i }=\frac{2}{9}(2 \hat{ i }-\hat{ j }+2 \hat{ k })+ t (2 \hat{ i }+2 \hat{ j }-\hat{ k }), t \in R$
$(3)$ $\overrightarrow{ r }=t(2 \hat{ i }+2 \hat{ j }-\hat{ k }), t \in R$
$(4)$ $\overrightarrow{ r }=\frac{2}{9}(4 \hat{ i }+\hat{ j }+\hat{ k })+ t (2 \hat{ i }+2 \hat{ j }-\hat{ k }), t \in R$