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Question Number 9109 by tawakalitu last updated on 18/Nov/16

if y = x^(log_a xy) find (dy/dx)

$$\mathrm{if}\:\:\:\mathrm{y}\:=\:\mathrm{x}^{\mathrm{log}_{\mathrm{a}} \mathrm{xy}} \\ $$$$\mathrm{find}\:\:\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$

Answered by mrW last updated on 19/Nov/16

y=x^(log_a xy) log_a y=(log_a x)(log_a xy) log_a y=(log_a x)(log_a x+log_a y) let u=log_a y, v=log_a x u=v(v+u) (1−v)u=v^2 u=(v^2 /(1−v)) log_a y=(v^2 /(1−v)) y=a^(v^2 /(1−v)) =a^(((log _a x)^2 )/(1−log _a x)) (dy/dv)=(ln a)a^(v^2 /(1−v)) (((2v)/(1−v))+(v^2 /((1−v)^2 )))=(ln a)(a^(v^2 /(1−v)) )(((v(2−v))/((1−v)^2 ))) (dv/dx)=(1/((ln a)x)) (dy/dx)=(dy/dv)∙(dv/dx)=(ln a)(a^(v^2 /(1−v)) )(((v(2−v))/((1−v)^2 )))((1/((ln a)x))) =(1/x)∙((v(2−v))/((1−v)^2 ))∙a^(v^2 /(1−v)) (dy/dx)=(((log _a x)(2−log _a x))/(x(1−log _a x)^2 ))∙a^(((log _a x)^2 )/(1−log _a x))

$${y}={x}^{{log}_{{a}} {xy}} \\ $$$${log}_{{a}} {y}=\left({log}_{{a}} {x}\right)\left({log}_{{a}} {xy}\right) \\ $$$${log}_{{a}} {y}=\left({log}_{{a}} {x}\right)\left({log}_{{a}} {x}+{log}_{{a}} {y}\right) \\ $$$${let}\:{u}={log}_{{a}} {y},\:{v}={log}_{{a}} {x} \\ $$$${u}={v}\left({v}+{u}\right) \\ $$$$\left(\mathrm{1}−{v}\right){u}={v}^{\mathrm{2}} \\ $$$${u}=\frac{{v}^{\mathrm{2}} }{\mathrm{1}−{v}} \\ $$$${log}_{{a}} {y}=\frac{{v}^{\mathrm{2}} }{\mathrm{1}−{v}} \\ $$$${y}={a}^{\frac{{v}^{\mathrm{2}} }{\mathrm{1}−{v}}} ={a}^{\frac{\left(\mathrm{log}\:_{{a}} {x}\right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{log}\:_{{a}} {x}}} \\ $$$$\frac{{dy}}{{dv}}=\left(\mathrm{ln}\:{a}\right){a}^{\frac{{v}^{\mathrm{2}} }{\mathrm{1}−{v}}} \left(\frac{\mathrm{2}{v}}{\mathrm{1}−{v}}+\frac{{v}^{\mathrm{2}} }{\left(\mathrm{1}−{v}\right)^{\mathrm{2}} }\right)=\left(\mathrm{ln}\:{a}\right)\left({a}^{\frac{{v}^{\mathrm{2}} }{\mathrm{1}−{v}}} \right)\left(\frac{{v}\left(\mathrm{2}−{v}\right)}{\left(\mathrm{1}−{v}\right)^{\mathrm{2}} }\right) \\ $$$$\frac{{dv}}{{dx}}=\frac{\mathrm{1}}{\left(\mathrm{ln}\:{a}\right){x}} \\ $$$$\frac{{dy}}{{dx}}=\frac{{dy}}{{dv}}\centerdot\frac{{dv}}{{dx}}=\left(\mathrm{ln}\:{a}\right)\left({a}^{\frac{{v}^{\mathrm{2}} }{\mathrm{1}−{v}}} \right)\left(\frac{{v}\left(\mathrm{2}−{v}\right)}{\left(\mathrm{1}−{v}\right)^{\mathrm{2}} }\right)\left(\frac{\mathrm{1}}{\left(\mathrm{ln}\:{a}\right){x}}\right) \\ $$$$=\frac{\mathrm{1}}{{x}}\centerdot\frac{{v}\left(\mathrm{2}−{v}\right)}{\left(\mathrm{1}−{v}\right)^{\mathrm{2}} }\centerdot{a}^{\frac{{v}^{\mathrm{2}} }{\mathrm{1}−{v}}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\left(\mathrm{log}\:_{{a}} {x}\right)\left(\mathrm{2}−\mathrm{log}\:_{{a}} {x}\right)}{{x}\left(\mathrm{1}−\mathrm{log}\:_{{a}} {x}\right)^{\mathrm{2}} }\centerdot{a}^{\frac{\left(\mathrm{log}\:_{{a}} {x}\right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{log}\:_{{a}} {x}}} \\ $$

Commented by tawakalitu last updated on 19/Nov/16

I really appreciate your effort sir. God bless you.

$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless} \\ $$$$\mathrm{you}. \\ $$

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