Question Number 82993 by aseer imad last updated on 26/Feb/20
Answered by TANMAY PANACEA last updated on 27/Feb/20
$$\left({x}+{y}\right)^{{x}+{y}} ={x}^{{y}} +{y}^{{x}} \\ $$$${a}={b}+{c} \\ $$$$\frac{{da}}{{dx}}=\frac{{db}}{{dx}}+\frac{{dc}}{{dx}} \\ $$$${now}\:\:\:{a}=\left({x}+{y}\right)^{{x}+{y}} \\ $$$${lna}=\left({x}+{y}\right){ln}\left({x}+{y}\right) \\ $$$$\frac{\mathrm{1}}{{a}}×\frac{{da}}{{dx}}=\left({x}+{y}\right)×\frac{\mathrm{1}}{{x}+{y}}\left(\mathrm{1}+\frac{{dy}}{{dx}}\right)+{ln}\left({x}+{y}\right)×\left(\mathrm{1}+\frac{{dy}}{{dx}}\right) \\ $$$$\frac{\mathrm{1}}{{a}}×\frac{{da}}{{dx}}=\left(\mathrm{1}+\frac{{dy}}{{dx}}\right)\left\{\mathrm{1}+{ln}\left({x}+{y}\right)\right\} \\ $$$$\frac{{da}}{{dx}}=\left({x}+{y}\right)^{{x}+{y}} ×\left(\mathrm{1}+\frac{{dy}}{{dx}}\right)\left\{\mathrm{1}+{ln}\left({x}+{y}\right)\right\} \\ $$$${b}={x}^{{y}} \\ $$$${lnb}={ylnx} \\ $$$$\frac{\mathrm{1}}{{b}}×\frac{{db}}{{dx}}=\frac{{y}}{{x}}+\frac{{dy}}{{dx}}{lnx} \\ $$$$\frac{{db}}{{dx}}={x}^{{y}} \left(\frac{{y}}{{x}}+\frac{{dy}}{{dx}}{lnx}\right) \\ $$$${c}={y}^{{x}} \\ $$$${lnc}={xlny} \\ $$$$\frac{\mathrm{1}}{{c}}\frac{{dc}}{{dx}}=\frac{{x}}{{y}}\frac{{dy}}{{dx}}+{lny} \\ $$$$\frac{{dc}}{{dx}}={y}^{{x}} \left(\frac{{x}}{{y}}\frac{{dy}}{{dx}}+{lny}\right) \\ $$$$\frac{{da}}{{dx}}=\frac{{db}}{{dx}}+\frac{{dc}}{{dx}} \\ $$$$\left({x}+{y}\right)^{{x}+{y}} \left(\mathrm{1}+\frac{{dy}}{{dx}}\right)\left\{\mathrm{1}+{ln}\left({x}+{y}\right)\right\}={x}^{{y}} \left(\frac{{y}}{{x}}+\frac{{dy}}{{dx}}{lnx}\right)+{y}^{{x}} \left(\frac{{x}}{{y}}\frac{{dy}}{{dx}}+{lny}\right) \\ $$$${now}\:{simplify}… \\ $$$$ \\ $$