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Question Number 207442 by York12 last updated on 15/May/24

$$\mathrm{If}y=f\left(x\right),\frac{d^{2}x}{{dy}^2}=e^{y+1},\mathrm{and}\mathrm{the}\mathrm{tangent}\mathrm{line}\mathrm{to}\mathrm{the}\mathrm{curve}\mathrm{of}\mathrm{the}\mathrm{function}f\left(x\right)\mathrm{on}\mathrm{the}\mathrm{point}$$$$(x_1,-1)\mathrm{is}\mathrm{paralel}\mathrm{to}\mathrm{the}\mathrm{straight}\mathrm{line}g\left(x\right)=x-3,\mathrm{then}\mathrm{find}{f}^{\prime }\left(x\right).$$

Commented by York12 last updated on 15/May/24

$$\mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{that}\mathrm{question}\mathrm{was}\mathrm{as}\mathrm{follows}\mathrm{but}\mathrm{I}\mathrm{do}\mathrm{not}\mathrm{understand}\mathrm{particular}\mathrm{things}$$$$\mathrm{he}\mathrm{said}\mathrm{since}\frac{d^{2}x}{{dy}^2}=e^{y+1}\stackrel{\mathrm{By}\mathrm{integrating}\mathrm{both}\mathrm{sides}}{\Rightarrow }\frac{d\left(x\right)}{d\left(y\right)}=e^{y+1}+c$$$$\mathrm{What}\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{understand}\mathrm{here}\mathrm{is}\mathrm{to}\mathrm{respect}\mathrm{to}\mathrm{what}$$$$\mathrm{he}\mathrm{integrated}\mathrm{both}\mathrm{sides}!$$$$\mathrm{I}\mathrm{know}\mathrm{when}\mathrm{I}\mathrm{have}\mathrm{something}\mathrm{like}\mathrm{for}\mathrm{example}$$$$3y^2\times \frac{d\left(y\right)}{d\left(x\right)}=2x+1\iff 3y^2\times d\left(y\right)=(2x+1)d\left(x\right)$$$$\mathrm{In}\mathrm{this}\mathrm{case}\mathrm{we}\mathrm{can}\mathrm{integrate}\mathrm{both}\mathrm{sides}$$$$\int 3y^{2}d\left(y\right)=\int (2x+1)d\left(x\right)\iff y^3=x^2+x+c$$$$\mathrm{I}\mathrm{do}\mathrm{not}\mathrm{understand}\mathrm{how}\mathrm{this}\mathrm{process}\mathrm{can}\mathrm{be}\mathrm{applied}\mathrm{to}\mathrm{the}\mathrm{problem}\mathrm{above}$$$$\mathrm{after}\mathrm{this}\mathrm{step}\mathrm{he}\mathrm{did}\mathrm{something}\mathrm{seems}\mathrm{wrong}$$$$\mathrm{he}\mathrm{said}\mathrm{that}\mathrm{since}\frac{d\left(y\right)}{d\left(x\right)}=\mathrm{The}\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{tangent}=1=\frac{d\left(x\right)}{d\left(y\right)}$$$$\mathrm{he}\mathrm{dealt}\mathrm{with}\mathrm{them}\mathrm{as}\mathrm{if}\mathrm{they}\mathrm{are}\mathrm{variables}\mathrm{divided}\mathrm{by}\mathrm{each}\mathrm{other}$$$$\mathrm{he}\mathrm{used}\mathrm{that}\mathrm{in}\frac{d\left(x\right)}{d\left(y\right)}=e^{y+1}+c\frac{d\left(x\right)}{d\left(y\right)}=e^{y+1}+c\Rightarrow 1={\mathrm{e}}^0+\mathrm{c}\Rightarrow \mathrm{c}=0$$$$\therefore \frac{d\left(x\right)}{d\left(y\right)}=e^{y+1},\mathrm{after}\mathrm{that}\mathrm{he}\mathrm{did}\mathrm{the}\mathrm{same}\mathrm{strange}\mathrm{thinh}$$$$\frac{d\left(y\right)}{d\left(x\right)}=\frac{1}{e^{y+1}}\Rightarrow \left(\mathrm{Which}\mathrm{is}\mathrm{strangely}\mathrm{true}\right)$$$$\mathrm{I}\mathrm{would}\mathrm{be}\mathrm{thanfull}\mathrm{for}\mathrm{anyone}\mathrm{who}\mathrm{can}\mathrm{explain}\mathrm{that},\mathrm{thanks}\mathrm{in}\mathrm{advance}$$$$$$

Commented by A5T last updated on 15/May/24

$$y=f\left(x\right),f\left(x_1\right)=-1;f^{{}^{\prime }}\left(x_1\right)=1$$$$\frac{d}{dy}\left(\frac{dx}{dy}\right)=e^{y+1}.$$$$Assumexhasbeenrepresentedasafunction$$$$ofy\Rightarrow x=f\left(y\right);thenyoucandifferentiate$$$$takingxasafunctionofy.$$$$x=f\left(y\right)\Rightarrow \frac{dx}{dy}=f^{{}^{\prime }}\left(y\right)\Rightarrow \frac{d}{dy}\left(\frac{dx}{dy}\right)=f^{{}^{″}}y=e^{y+1}$$$$\int \int f^{″}y=e^{y+1}+cy+k=f\left(y\right)$$$$\Rightarrow x=f\left(y\right)=e^{y+1}+cy+k\Rightarrow \frac{dx}{dy}=e^{y+1}+c$$$$\Rightarrow \frac{dy}{dx}=\frac{1}{e^{y+1}+c};$$$$\frac{dy}{dx}=1wheny=-1\Rightarrow \frac{1}{1+c}=1\Rightarrow c=0$$$$\Rightarrow x=e^{y+1}+k\Rightarrow y=ln(x-k)-1;k=x_1-1$$$$$$

Commented by York12 last updated on 15/May/24

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

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

$$\frac{d^{2}x}{{dy}^2}=\frac{d}{dy}\left(\frac{dx}{dy}\right)=e^{y+1}$$$$\Rightarrow \frac{dx}{dy}=e^{y+1}+C_1$$$$\Rightarrow x=e^{y+1}+C_{1}y+C_2$$$$at(x_1,-1):$$$$\frac{dx}{dy}{\mid }_{y=-1}=e^{-1+1}+C_1\stackrel{!}{=}1\Rightarrow C_1=0$$$$x_1=e^{-1+1}+C_2\Rightarrow C_2=x_1-1$$$$\Rightarrow x=e^{y+1}+x_1-1$$$$\Rightarrow y=\ln (x-x_1+1)-1=f\left(x\right)$$$$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{x-x_1+1}$$

Commented by York12 last updated on 15/May/24

$$\mathrm{Really},\mathrm{Really}\mathrm{appreciate}\mathrm{it}$$

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