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Question Number 207450 by York12 last updated on 15/May/24

$$\mathrm{Find}\mathrm{the}\mathrm{relation}\mathrm{between}m\mathrm{and}n\mathrm{for}\mathrm{which}\mathrm{the}\mathrm{following}\mathrm{holds}$$$$\frac{d\left(y\right)}{d\left(x\right)}{\mid }_{x=n}={\left(\frac{d\left(x\right)}{d\left(y\right)}{\mid }_{y=m}\right)}^{-1}$$

Answered by mr W last updated on 15/May/24

Commented by mr W last updated on 15/May/24

$$\frac{dy}{dx}{\mid }_{x=n}=\tan \theta$$$$\frac{dx}{dy}{\mid }_{y=m}=\tan \varphi$$$${\left(\frac{dx}{dy}{\mid }_{y=m}\right)}^{-1}=\frac{1}{\tan \varphi }=\tan \phi$$$$\frac{dy}{dx}{\mid }_{x=n}={\left(\frac{dx}{dy}{\mid }_{y=m}\right)}^{-1}$$$$\Rightarrow \tan \theta =\tan \phi$$$$\Rightarrow \theta =\phi$$$$ify=f\left(x\right)isastrictlymonotonic$$$$function,thenPandQmustbe$$$$thesamepoint,i.e.m=f\left(n\right).$$$$ify=f\left(x\right)isnotastrictlymonotonic$$$$function,thentherelationbetween$$$$nundmisnotunique.$$

Commented by York12 last updated on 15/May/24

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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