Question-221721

Question Number 221721 by ahmed2025 last updated on 09/Jun/25

Answered by wewji12 last updated on 09/Jun/25

(dy/(dln(t)))=((dy/dt)/((d{ln(t)})/dt)) (Warning (dy/dx) is NOT Fraction!!!) ((((d )/dt){t^2 ln(t)})/(((d )/dt){ln(t)}))=((2tln(t)+t)/(1/t))=t^2 +2t^2 ln(t)=t^2 (1+2ln(t)) ((dy(t))/(d{ln(t)}))=t^2 (1+2ln(t))

$$\frac{\mathrm{d}{y}}{\mathrm{dln}\left({t}\right)}=\frac{\frac{\mathrm{d}{y}}{\mathrm{d}{t}}}{\frac{\mathrm{d}\left\{\mathrm{ln}\left({t}\right)\right\}}{\mathrm{d}{t}}}\:\left(\mathrm{Warning}\:\:\frac{\mathrm{d}{y}}{\mathrm{d}{x}}\:\mathrm{is}\:\mathrm{NOT}\:\mathrm{Fraction}!!!\right) \\ $$$$\frac{\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\left\{{t}^{\mathrm{2}} \mathrm{ln}\left({t}\right)\right\}}{\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\left\{\mathrm{ln}\left({t}\right)\right\}}=\frac{\mathrm{2}{t}\mathrm{ln}\left({t}\right)+{t}}{\frac{\mathrm{1}}{{t}}}={t}^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{2}} \mathrm{ln}\left({t}\right)={t}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2ln}\left({t}\right)\right) \\ $$$$\frac{\mathrm{d}{y}\left({t}\right)}{\mathrm{d}\left\{\mathrm{ln}\left({t}\right)\right\}}={t}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2ln}\left({t}\right)\right) \\ $$

Commented by wewji12 last updated on 09/Jun/25

If you′re wondering why ′′(dy/dx)′′ isn′t a fraction I encourage you to look up the exterior-derivative and differantial form

$$ \\ $$$$\mathrm{If}\:\mathrm{you}'\mathrm{re}\:\mathrm{wondering}\:\mathrm{why}\:''\frac{\mathrm{d}{y}}{\mathrm{d}{x}}''\:\mathrm{isn}'\mathrm{t}\:\mathrm{a}\:\mathrm{fraction} \\ $$$$\mathrm{I}\:\mathrm{encourage}\:\mathrm{you}\:\mathrm{to}\:\mathrm{look}\:\mathrm{up}\:\mathrm{the}\:\mathrm{exterior}-\mathrm{derivative} \\ $$$$\mathrm{and}\:\mathrm{differantial}\:\mathrm{form} \\ $$

Commented by fantastic last updated on 09/Jun/25

$${that}\:{warning}\:{seems}\:{funny}\:{to}\:\:{me} \\ $$

Commented by wewji12 last updated on 09/Jun/25

heh.... And if you are wondering why a notation like ((dy(t))/(d{ln(t)})) is allowed in mathematics as you asked I would like you to refer to the concept called Radon−Nikodym derivaive (dν/dμ) among the concepts used in measure theory.

$$\mathrm{heh}…. \\ $$$$ \\ $$$$\mathrm{And}\:\mathrm{if}\:\mathrm{you}\:\mathrm{are}\:\mathrm{wondering}\:\mathrm{why}\:\mathrm{a}\:\mathrm{notation}\:\mathrm{like}\:\frac{\mathrm{d}{y}\left({t}\right)}{\mathrm{d}\left\{\mathrm{ln}\left({t}\right)\right\}} \\ $$$$\mathrm{is}\:\mathrm{allowed}\:\mathrm{in}\:\mathrm{mathematics}\:\mathrm{as}\:\mathrm{you}\:\mathrm{asked} \\ $$$$\mathrm{I}\:\mathrm{would}\:\mathrm{like}\:\mathrm{you}\:\mathrm{to}\:\mathrm{refer}\:\mathrm{to}\:\mathrm{the}\:\mathrm{concept}\:\mathrm{called} \\ $$$$\mathrm{Radon}−\mathrm{Nikodym}\:\mathrm{derivaive}\:\frac{\mathrm{d}\nu}{\mathrm{d}\mu} \\ $$$$\mathrm{among}\:\mathrm{the}\:\mathrm{concepts}\:\mathrm{used}\:\mathrm{in}\:\mathrm{measure}\:\mathrm{theory}. \\ $$

Commented by wewji12 last updated on 09/Jun/25

hmmm...Let′s just say.. if measurable space (X,𝚺) is defined and measure-μ,ν satisfying μ(℧)=1 and 0≤ν{F}≤μ{F} for all F∈F and.... Satisfy ν(F)=∫_F h dμ and exist F-measurable function for h;℧→R .. and we can renote h as h=(dν/dμ) we call it Radon-Nikodym derivate

$${hmmm}…\mathrm{Let}'\mathrm{s}\:\mathrm{just}\:\mathrm{say}.. \\ $$$$\mathrm{if}\:\mathrm{measurable}\:\mathrm{space}\:\left({X},\boldsymbol{\Sigma}\right)\:\mathrm{is}\:\mathrm{defined}\:\mathrm{and} \\ $$$$\mathrm{measure}-\mu,\nu\:\mathrm{satisfying}\:\mu\left(\mho\right)=\mathrm{1}\:\mathrm{and} \\ $$$$\mathrm{0}\leq\nu\left\{\mathcal{F}\right\}\leq\mu\left\{\mathcal{F}\right\}\:\mathrm{for}\:\mathrm{all}\:{F}\in\mathcal{F}\:\mathrm{and}…. \\ $$$$\mathrm{Satisfy}\:\nu\left({F}\right)=\int_{{F}} \:{h}\:\mathrm{d}\mu\:\mathrm{and}\: \\ $$$$\mathrm{exist}\:{F}-\mathrm{measurable}\:\mathrm{function}\:\mathrm{for}\:{h};\mho\rightarrow\mathbb{R}\:.. \\ $$$$\mathrm{and}\:\mathrm{we}\:\mathrm{can}\:\mathrm{renote}\:{h}\:\mathrm{as}\:{h}=\frac{\mathrm{d}\nu}{\mathrm{d}\mu} \\ $$$$\mathrm{we}\:\mathrm{call}\:\mathrm{it}\:\mathrm{Radon}-\mathrm{Nikodym}\:\mathrm{derivate} \\ $$

Answered by mr W last updated on 09/Jun/25

let t=ln x ⇒x=e^t y=e^(2t) t (dy/(d(ln x)))=(dy/dt)=e^(2t) +2e^(2t) t =e^(2t) (1+2t)=x^2 (1+2ln x)

$${let}\:{t}=\mathrm{ln}\:{x} \\ $$$$\Rightarrow{x}={e}^{{t}} \\ $$$${y}={e}^{\mathrm{2}{t}} {t} \\ $$$$\frac{{dy}}{{d}\left(\mathrm{ln}\:{x}\right)}=\frac{{dy}}{{dt}}={e}^{\mathrm{2}{t}} +\mathrm{2}{e}^{\mathrm{2}{t}} {t} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={e}^{\mathrm{2}{t}} \left(\mathrm{1}+\mathrm{2}{t}\right)={x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2ln}\:{x}\right) \\ $$

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