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Question Number 38181 by rahul 19 last updated on 22/Jun/18
If y= x^((lnx)^(ln(lnx)) ) then (dy/dx) = ?
$$\mathrm{If}\:\mathrm{y}=\:\:{x}^{\left({lnx}\right)^{{ln}\left({lnx}\right)} } \:{then}\:\frac{{dy}}{{dx}}\:=\:? \\ $$
Commented by rahul 19 last updated on 22/Jun/18
I ′ve done by taking log and I′m getting (y/x)(lnx)^(ln(lnx)) (ln(lnx)+1). Someone pls verify.
$$\mathrm{I}\:'\mathrm{ve}\:\mathrm{done}\:\mathrm{by}\:\mathrm{taking}\:\mathrm{log}\:\mathrm{and}\:\mathrm{I}'\mathrm{m}\:\mathrm{getting} \\ $$$$\frac{\mathrm{y}}{{x}}\left({lnx}\right)^{{ln}\left({lnx}\right)} \left({ln}\left({lnx}\right)+\mathrm{1}\right). \\ $$$${S}\mathrm{omeone}\:\mathrm{pls}\:\mathrm{verify}. \\ $$
Commented by MJS last updated on 23/Jun/18
it′s almost right, but the last factor must be (2ln(ln x)+1)
$$\mathrm{it}'\mathrm{s}\:\mathrm{almost}\:\mathrm{right},\:\mathrm{but}\:\mathrm{the}\:\mathrm{last}\:\mathrm{factor}\:\mathrm{must}\:\mathrm{be} \\ $$$$\:\left(\mathrm{2ln}\left(\mathrm{ln}\:{x}\right)+\mathrm{1}\right) \\ $$
Commented by abdo mathsup 649 cc last updated on 23/Jun/18
let put a(x)=(ln(x))^(ln(lnx)) ⇒y(x)= x^(a(x)) =e^(a(x)ln(x)) ⇒(dy/dx)(x)={xa(x)}^′ y(x) =(a(x) +xa^′ (x)) y(x) but a(x)=e^(ln(ln(x)ln(lnx)) =e^({ln(lnx)}^2 ⇒) a^′ (x)= 2(ln(lnx))(ln(lnx))^′ a(x) =(2/(xln(x)))ln(ln(x))a(x) ⇒ y^′ (x)=a(x) (1+x (2/(xln(x)))ln(ln(x)))y(x) =a(x)y(x)( 1+ ((2ln(ln(x)))/(ln(x)))).
$${let}\:{put}\:{a}\left({x}\right)=\left({ln}\left({x}\right)\right)^{{ln}\left({lnx}\right)} \:\Rightarrow{y}\left({x}\right)=\:{x}^{{a}\left({x}\right)} \\ $$$$={e}^{{a}\left({x}\right){ln}\left({x}\right)} \:\Rightarrow\frac{{dy}}{{dx}}\left({x}\right)=\left\{{xa}\left({x}\right)\right\}^{'} \:{y}\left({x}\right) \\ $$$$=\left({a}\left({x}\right)\:+{xa}^{'} \left({x}\right)\right)\:{y}\left({x}\right)\:{but} \\ $$$${a}\left({x}\right)={e}^{{ln}\left({ln}\left({x}\right){ln}\left({lnx}\right)\right.} ={e}^{\left\{{ln}\left({lnx}\right)\right\}^{\mathrm{2}} \:\Rightarrow} \\ $$$${a}^{'} \left({x}\right)=\:\mathrm{2}\left({ln}\left({lnx}\right)\right)\left({ln}\left({lnx}\right)\right)^{'} \:{a}\left({x}\right) \\ $$$$=\frac{\mathrm{2}}{{xln}\left({x}\right)}{ln}\left({ln}\left({x}\right)\right){a}\left({x}\right)\:\Rightarrow \\ $$$${y}^{'} \left({x}\right)={a}\left({x}\right)\:\left(\mathrm{1}+{x}\:\frac{\mathrm{2}}{{xln}\left({x}\right)}{ln}\left({ln}\left({x}\right)\right)\right){y}\left({x}\right) \\ $$$$={a}\left({x}\right){y}\left({x}\right)\left(\:\mathrm{1}+\:\frac{\mathrm{2}{ln}\left({ln}\left({x}\right)\right)}{{ln}\left({x}\right)}\right). \\ $$
Answered by MJS last updated on 23/Jun/18
(d/dx)[x^(f(x)) ]=x^(f(x)) ×(((f(x))/x)+(d/dx)[f(x)]×ln x) f(x)=(ln x)^(g(x)) (d/dx)[f(x)]=(ln x)^(g(x)) ×(((g(x))/(xln x))+(d/dx)[g(x)]×ln(ln x)) g(x)=ln(ln x) (d/dx)[g(x)]=(1/(xln x)) (d/dx)[f(x)]=(ln x)^(ln(ln x)) ×(((ln(ln x))/(xln x))+(1/(xln x))×ln(ln x))= =2((ln(ln x))/(xln x))(ln x)^(ln(ln x)) (d/dx)[x^(f(x)) ]=x^((ln x)^(ln(ln x)) ) ×((((ln x)^(ln(ln x)) )/x)+2((ln(ln x))/(xln x))(ln x)^(ln(ln x)) ×ln x)= =(1+2ln(ln x))(ln x)^(ln(ln x)) x^((ln x)^(ln(ln x)) −1)
$$\frac{{d}}{{dx}}\left[{x}^{{f}\left({x}\right)} \right]={x}^{{f}\left({x}\right)} ×\left(\frac{{f}\left({x}\right)}{{x}}+\frac{{d}}{{dx}}\left[{f}\left({x}\right)\right]×\mathrm{ln}\:{x}\right) \\ $$$${f}\left({x}\right)=\left(\mathrm{ln}\:{x}\right)^{{g}\left({x}\right)} \\ $$$$\frac{{d}}{{dx}}\left[{f}\left({x}\right)\right]=\left(\mathrm{ln}\:{x}\right)^{{g}\left({x}\right)} ×\left(\frac{{g}\left({x}\right)}{{x}\mathrm{ln}\:{x}}+\frac{{d}}{{dx}}\left[{g}\left({x}\right)\right]×\mathrm{ln}\left(\mathrm{ln}\:{x}\right)\right) \\ $$$${g}\left({x}\right)=\mathrm{ln}\left(\mathrm{ln}\:{x}\right) \\ $$$$\frac{{d}}{{dx}}\left[{g}\left({x}\right)\right]=\frac{\mathrm{1}}{{x}\mathrm{ln}\:{x}} \\ $$$$ \\ $$$$\frac{{d}}{{dx}}\left[{f}\left({x}\right)\right]=\left(\mathrm{ln}\:{x}\right)^{\mathrm{ln}\left(\mathrm{ln}\:{x}\right)} ×\left(\frac{\mathrm{ln}\left(\mathrm{ln}\:{x}\right)}{{x}\mathrm{ln}\:{x}}+\frac{\mathrm{1}}{{x}\mathrm{ln}\:{x}}×\mathrm{ln}\left(\mathrm{ln}\:{x}\right)\right)= \\ $$$$\:\:\:\:\:=\mathrm{2}\frac{\mathrm{ln}\left(\mathrm{ln}\:{x}\right)}{{x}\mathrm{ln}\:{x}}\left(\mathrm{ln}\:{x}\right)^{\mathrm{ln}\left(\mathrm{ln}\:{x}\right)} \\ $$$$\frac{{d}}{{dx}}\left[{x}^{{f}\left({x}\right)} \right]={x}^{\left(\mathrm{ln}\:{x}\right)^{\mathrm{ln}\left(\mathrm{ln}\:{x}\right)} } ×\left(\frac{\left(\mathrm{ln}\:{x}\right)^{\mathrm{ln}\left(\mathrm{ln}\:{x}\right)} }{{x}}+\mathrm{2}\frac{\mathrm{ln}\left(\mathrm{ln}\:{x}\right)}{{x}\mathrm{ln}\:{x}}\left(\mathrm{ln}\:{x}\right)^{\mathrm{ln}\left(\mathrm{ln}\:{x}\right)} ×\mathrm{ln}\:{x}\right)= \\ $$$$\:\:\:\:\:=\left(\mathrm{1}+\mathrm{2ln}\left(\mathrm{ln}\:{x}\right)\right)\left(\mathrm{ln}\:{x}\right)^{\mathrm{ln}\left(\mathrm{ln}\:{x}\right)} {x}^{\left(\mathrm{ln}\:{x}\right)^{\mathrm{ln}\left(\mathrm{ln}\:{x}\right)} −\mathrm{1}} \\ $$
Answered by rahul 19 last updated on 23/Jun/18
lny = (lnx)^(ln(lnx)) (lnx) ⇒ ln y = ( lnx )^((ln(lnx)+1)) ⇒ (dy/dx)×(1/y) =( ln(lnx)+1)(lnx)^((ln(lnx))) ×(1/x) ⇒ (dy/dx) = (y/x) (1.ln(lnx)+1)(lnx)^((ln(lnx))) . Where am i wrong ?? Sir MJS ??
$${lny}\:=\:\left({lnx}\right)^{{ln}\left({lnx}\right)} \left({lnx}\right) \\ $$$$\Rightarrow\:\mathrm{ln}\:\mathrm{y}\:=\:\left(\:{lnx}\:\right)\:^{\left({ln}\left({lnx}\right)+\mathrm{1}\right)} \\ $$$$\Rightarrow\:\frac{\mathrm{dy}}{\mathrm{d}{x}}×\frac{\mathrm{1}}{{y}}\:=\left(\:{ln}\left({lnx}\right)+\mathrm{1}\right)\left({lnx}\right)^{\left({ln}\left({lnx}\right)\right)} ×\frac{\mathrm{1}}{{x}} \\ $$$$\Rightarrow\:\frac{\mathrm{dy}}{\mathrm{d}{x}}\:=\:\frac{{y}}{{x}}\:\left(\mathrm{1}.{ln}\left({lnx}\right)+\mathrm{1}\right)\left({lnx}\right)^{\left({ln}\left({lnx}\right)\right)} . \\ $$$${W}\mathrm{here}\:\mathrm{am}\:\mathrm{i}\:\mathrm{wrong}\:?? \\ $$$$\mathrm{Sir}\:\mathrm{MJS}\:?? \\ $$
Commented by MJS last updated on 23/Jun/18
please post the formula you′re using, I don′t understand how you get to this: (dy/dx)×(1/y) =( ln(lnx)+1)(lnx)^((ln(lnx))) ×(1/x)
$$\mathrm{please}\:\mathrm{post}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{you}'\mathrm{re}\:\mathrm{using},\:\mathrm{I}\:\mathrm{don}'\mathrm{t} \\ $$$$\mathrm{understand}\:\mathrm{how}\:\mathrm{you}\:\mathrm{get}\:\mathrm{to}\:\mathrm{this}: \\ $$$$\frac{\mathrm{dy}}{\mathrm{d}{x}}×\frac{\mathrm{1}}{{y}}\:=\left(\:{ln}\left({lnx}\right)+\mathrm{1}\right)\left({lnx}\right)^{\left({ln}\left({lnx}\right)\right)} ×\frac{\mathrm{1}}{{x}} \\ $$
Commented by rahul 19 last updated on 24/Jun/18
Ok i got my mistake ! :) i did like differentiate x^x then i simply wrote x.x^(x−1) .1 which is absolutely wrong!
$$\left.\mathrm{Ok}\:\mathrm{i}\:\mathrm{got}\:\mathrm{my}\:\mathrm{mistake}\:!\::\right) \\ $$$$\mathrm{i}\:\mathrm{did}\:\mathrm{like}\:\mathrm{differentiate}\:{x}^{{x}} \\ $$$${then}\:{i}\:{simply}\:{wrote}\:{x}.{x}^{{x}−\mathrm{1}} .\mathrm{1}\:{which} \\ $$$${is}\:{absolutely}\:{wrong}! \\ $$
Commented by MJS last updated on 24/Jun/18
good! learning by doing and learning by mistakes :−)
$$\mathrm{good}! \\ $$$$\mathrm{learning}\:\mathrm{by}\:\mathrm{doing}\:\mathrm{and}\:\mathrm{learning}\:\mathrm{by}\:\mathrm{mistakes} \\ $$$$\left.:−\right) \\ $$

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