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

If y=f(x), (d^2 x/dy^2 )=e^(y+1) , and the tangent line to the curve of the function f(x) on the point (x_1 ,−1) is paralel to the straight line g(x)=x−3, then find f′(x).

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

Commented by York12 last updated on 15/May/24

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

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

Commented by A5T last updated on 15/May/24

y=f(x), f(x_1 )=−1 ;f^′ (x_1 )=1 (d/dy)((dx/dy))=e^(y+1) . Assume x has been represented as a function of y⇒x=f(y); then you can differentiate taking x as a function of y. x=f(y)⇒(dx/dy)=f^′ (y)⇒(d/dy)((dx/dy))=f^(′′) y=e^(y+1) ∫∫f′′y=e^(y+1) +cy+k=f(y) ⇒x=f(y)=e^(y+1) +cy+k⇒(dx/dy)=e^(y+1) +c ⇒(dy/dx)=(1/(e^(y+1) +c)); (dy/dx)=1 when y=−1⇒(1/(1+c))=1⇒c=0 ⇒x=e^(y+1) +k⇒y=ln(x−k)−1; k=x_1 −1

$${y}={f}\left({x}\right),\:{f}\left({x}_{\mathrm{1}} \right)=−\mathrm{1}\:;{f}^{'} \left({x}_{\mathrm{1}} \right)=\mathrm{1} \\ $$$$\frac{{d}}{{dy}}\left(\frac{{dx}}{{dy}}\right)={e}^{{y}+\mathrm{1}} . \\ $$$${Assume}\:{x}\:{has}\:{been}\:{represented}\:{as}\:{a}\:{function}\: \\ $$$${of}\:{y}\Rightarrow{x}={f}\left({y}\right);\:{then}\:{you}\:{can}\:{differentiate} \\ $$$${taking}\:{x}\:{as}\:{a}\:{function}\:{of}\:{y}. \\ $$$${x}={f}\left({y}\right)\Rightarrow\frac{{dx}}{{dy}}={f}^{'} \left({y}\right)\Rightarrow\frac{{d}}{{dy}}\left(\frac{{dx}}{{dy}}\right)={f}^{''} {y}={e}^{{y}+\mathrm{1}} \\ $$$$\int\int{f}''{y}={e}^{{y}+\mathrm{1}} +{cy}+{k}={f}\left({y}\right) \\ $$$$\Rightarrow{x}={f}\left({y}\right)={e}^{{y}+\mathrm{1}} +{cy}+{k}\Rightarrow\frac{{dx}}{{dy}}={e}^{{y}+\mathrm{1}} +{c} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{{e}^{{y}+\mathrm{1}} +{c}}; \\ $$$$\frac{{dy}}{{dx}}=\mathrm{1}\:{when}\:{y}=−\mathrm{1}\Rightarrow\frac{\mathrm{1}}{\mathrm{1}+{c}}=\mathrm{1}\Rightarrow{c}=\mathrm{0} \\ $$$$\Rightarrow{x}={e}^{{y}+\mathrm{1}} +{k}\Rightarrow{y}={ln}\left({x}−{k}\right)−\mathrm{1};\:{k}={x}_{\mathrm{1}} −\mathrm{1} \\ $$$$ \\ $$

Commented by York12 last updated on 15/May/24

thank you sir

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

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

(d^2 x/dy^2 )=(d/dy)((dx/dy))=e^(y+1) ⇒(dx/dy)=e^(y+1) +C_1 ⇒x=e^(y+1) +C_1 y+C_2 at (x_1 , −1): (dx/dy)∣_(y=−1) =e^(−1+1) +C_1 =^! 1 ⇒C_1 =0 x_1 =e^(−1+1) +C_2 ⇒C_2 =x_1 −1 ⇒x=e^(y+1) +x_1 −1 ⇒y=ln (x−x_1 +1)−1=f(x) ⇒f′(x)=(1/(x−x_1 +1))

$$\frac{{d}^{\mathrm{2}} {x}}{{dy}^{\mathrm{2}} }=\frac{{d}}{{dy}}\left(\frac{{dx}}{{dy}}\right)={e}^{{y}+\mathrm{1}} \\ $$$$\Rightarrow\frac{{dx}}{{dy}}={e}^{{y}+\mathrm{1}} +{C}_{\mathrm{1}} \\ $$$$\Rightarrow{x}={e}^{{y}+\mathrm{1}} +{C}_{\mathrm{1}} {y}+{C}_{\mathrm{2}} \\ $$$${at}\:\left({x}_{\mathrm{1}} ,\:−\mathrm{1}\right): \\ $$$$\frac{{dx}}{{dy}}\mid_{{y}=−\mathrm{1}} ={e}^{−\mathrm{1}+\mathrm{1}} +{C}_{\mathrm{1}} \overset{!} {=}\mathrm{1}\:\Rightarrow{C}_{\mathrm{1}} =\mathrm{0} \\ $$$${x}_{\mathrm{1}} ={e}^{−\mathrm{1}+\mathrm{1}} +{C}_{\mathrm{2}} \:\Rightarrow{C}_{\mathrm{2}} ={x}_{\mathrm{1}} −\mathrm{1} \\ $$$$\Rightarrow{x}={e}^{{y}+\mathrm{1}} +{x}_{\mathrm{1}} −\mathrm{1} \\ $$$$\Rightarrow{y}=\mathrm{ln}\:\left({x}−{x}_{\mathrm{1}} +\mathrm{1}\right)−\mathrm{1}={f}\left({x}\right) \\ $$$$\Rightarrow{f}'\left({x}\right)=\frac{\mathrm{1}}{{x}−{x}_{\mathrm{1}} +\mathrm{1}} \\ $$

Commented by York12 last updated on 15/May/24

Really , Really appreciate it

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

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