if-y-x-log-a-xy-find-dy-dx-

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

if y = x^{\log_{a} xy}

find \frac{dy}{dx}

Answered by mrW last updated on 19/Nov/16

y = x^{{l o g}_{a} x y}

{l o g}_{a} y = ({l o g}_{a} x) ({l o g}_{a} x y)

{l o g}_{a} y = ({l o g}_{a} x) ({l o g}_{a} x + {l o g}_{a} y)

l e t u = {l o g}_{a} y, v = {l o g}_{a} x

u = v (v + u)

(1 - v) u = v^{2}

u = \frac{v^{2}}{1 - v}

{l o g}_{a} y = \frac{v^{2}}{1 - v}

y = a^{\frac{v^{2}}{1 - v}} = a^{\frac{{(\log_{a} x)}^{2}}{1 - \log_{a} x}}

\frac{d y}{d v} = (\ln a) a^{\frac{v^{2}}{1 - v}} (\frac{2 v}{1 - v} + \frac{v^{2}}{{(1 - v)}^{2}}) = (\ln a) (a^{\frac{v^{2}}{1 - v}}) (\frac{v (2 - v)}{{(1 - v)}^{2}})

\frac{d v}{d x} = \frac{1}{(\ln a) x}

\frac{d y}{d x} = \frac{d y}{d v} \cdot \frac{d v}{d x} = (\ln a) (a^{\frac{v^{2}}{1 - v}}) (\frac{v (2 - v)}{{(1 - v)}^{2}}) (\frac{1}{(\ln a) x})

= \frac{1}{x} \cdot \frac{v (2 - v)}{{(1 - v)}^{2}} \cdot a^{\frac{v^{2}}{1 - v}}

\frac{d y}{d x} = \frac{(\log_{a} x) (2 - \log_{a} x)}{x {(1 - \log_{a} x)}^{2}} \cdot a^{\frac{{(\log_{a} x)}^{2}}{1 - \log_{a} x}}

Commented by tawakalitu last updated on 19/Nov/16

I really appreciate your effort sir . God bless

you .