Question Number 9109 by tawakalitu last updated on 18/Nov/16
$$\mathrm{if}\:\:\:\mathrm{y}\:=\:\mathrm{x}^{\mathrm{log}_{\mathrm{a}} \mathrm{xy}} \\ $$$$\mathrm{find}\:\:\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$
Answered by mrW last updated on 19/Nov/16
$${y}={x}^{{log}_{{a}} {xy}} \\ $$$${log}_{{a}} {y}=\left({log}_{{a}} {x}\right)\left({log}_{{a}} {xy}\right) \\ $$$${log}_{{a}} {y}=\left({log}_{{a}} {x}\right)\left({log}_{{a}} {x}+{log}_{{a}} {y}\right) \\ $$$${let}\:{u}={log}_{{a}} {y},\:{v}={log}_{{a}} {x} \\ $$$${u}={v}\left({v}+{u}\right) \\ $$$$\left(\mathrm{1}−{v}\right){u}={v}^{\mathrm{2}} \\ $$$${u}=\frac{{v}^{\mathrm{2}} }{\mathrm{1}−{v}} \\ $$$${log}_{{a}} {y}=\frac{{v}^{\mathrm{2}} }{\mathrm{1}−{v}} \\ $$$${y}={a}^{\frac{{v}^{\mathrm{2}} }{\mathrm{1}−{v}}} ={a}^{\frac{\left(\mathrm{log}\:_{{a}} {x}\right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{log}\:_{{a}} {x}}} \\ $$$$\frac{{dy}}{{dv}}=\left(\mathrm{ln}\:{a}\right){a}^{\frac{{v}^{\mathrm{2}} }{\mathrm{1}−{v}}} \left(\frac{\mathrm{2}{v}}{\mathrm{1}−{v}}+\frac{{v}^{\mathrm{2}} }{\left(\mathrm{1}−{v}\right)^{\mathrm{2}} }\right)=\left(\mathrm{ln}\:{a}\right)\left({a}^{\frac{{v}^{\mathrm{2}} }{\mathrm{1}−{v}}} \right)\left(\frac{{v}\left(\mathrm{2}−{v}\right)}{\left(\mathrm{1}−{v}\right)^{\mathrm{2}} }\right) \\ $$$$\frac{{dv}}{{dx}}=\frac{\mathrm{1}}{\left(\mathrm{ln}\:{a}\right){x}} \\ $$$$\frac{{dy}}{{dx}}=\frac{{dy}}{{dv}}\centerdot\frac{{dv}}{{dx}}=\left(\mathrm{ln}\:{a}\right)\left({a}^{\frac{{v}^{\mathrm{2}} }{\mathrm{1}−{v}}} \right)\left(\frac{{v}\left(\mathrm{2}−{v}\right)}{\left(\mathrm{1}−{v}\right)^{\mathrm{2}} }\right)\left(\frac{\mathrm{1}}{\left(\mathrm{ln}\:{a}\right){x}}\right) \\ $$$$=\frac{\mathrm{1}}{{x}}\centerdot\frac{{v}\left(\mathrm{2}−{v}\right)}{\left(\mathrm{1}−{v}\right)^{\mathrm{2}} }\centerdot{a}^{\frac{{v}^{\mathrm{2}} }{\mathrm{1}−{v}}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\left(\mathrm{log}\:_{{a}} {x}\right)\left(\mathrm{2}−\mathrm{log}\:_{{a}} {x}\right)}{{x}\left(\mathrm{1}−\mathrm{log}\:_{{a}} {x}\right)^{\mathrm{2}} }\centerdot{a}^{\frac{\left(\mathrm{log}\:_{{a}} {x}\right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{log}\:_{{a}} {x}}} \\ $$
Commented by tawakalitu last updated on 19/Nov/16
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless} \\ $$$$\mathrm{you}. \\ $$