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-

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

If y = f (x), \frac{d^{2} x}{{d y}^{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) .

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 \frac{d^{2} x}{{d y}^{2}} = e^{y + 1} \overset{By integrating both sides}{\Rightarrow} \frac{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

3 y^{2} \times \frac{d (y)}{d (x)} = 2 x + 1 \Leftrightarrow 3 y^{2} \times d (y) = (2 x + 1) d (x)

In this case we can integrate both sides

\int 3 y^{2} d (y) = \int (2 x + 1) d (x) \Leftrightarrow 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 \frac{d (y)}{d (x)} = The slope of the tangent = 1 = \frac{d (x)}{d (y)}

he dealt with them as if they are variables divided by each other

he used that in \frac{d (x)}{d (y)} = e^{y + 1} + c \frac{d (x)}{d (y)} = e^{y + 1} + c \Rightarrow 1 = e^{0} + c \Rightarrow c = 0

∴ \frac{d (x)}{d (y)} = e^{y + 1}, after that he did the same strange thinh

\frac{d (y)}{d (x)} = \frac{1}{e^{y + 1}} \Rightarrow (Which is strangely true)

I would be thanfull for anyone who can explain that, thanks in advance

Commented by A5T last updated on 15/May/24

y = f (x), f (x_{1}) = - 1; f^{^{'}} (x_{1}) = 1

\frac{d}{d y} (\frac{d x}{d y}) = e^{y + 1} .

A s s u m e x h a s b e e n r e p r e s e n t e d a s a f u n c t i o n

o f y \Rightarrow x = f (y); t h e n y o u c a n d i f f e r e n t i a t e

t a k i n g x a s a f u n c t i o n o f y .

x = f (y) \Rightarrow \frac{d x}{d y} = f^{^{'}} (y) \Rightarrow \frac{d}{d y} (\frac{d x}{d y}) = f^{^{″}} y = e^{y + 1}

\int \int f^{″} y = e^{y + 1} + c y + k = f (y)

\Rightarrow x = f (y) = e^{y + 1} + c y + k \Rightarrow \frac{d x}{d y} = e^{y + 1} + c

\Rightarrow \frac{d y}{d x} = \frac{1}{e^{y + 1} + c};

\frac{d y}{d x} = 1 w h e n y = - 1 \Rightarrow \frac{1}{1 + c} = 1 \Rightarrow c = 0

\Rightarrow x = e^{y + 1} + k \Rightarrow y = l n (x - k) - 1; k = x_{1} - 1

Commented by York12 last updated on 15/May/24

thank you sir

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

\frac{d^{2} x}{{d y}^{2}} = \frac{d}{d y} (\frac{d x}{d y}) = e^{y + 1}

\Rightarrow \frac{d x}{d y} = e^{y + 1} + C_{1}

\Rightarrow x = e^{y + 1} + C_{1} y + C_{2}

a t (x_{1}, - 1) :

\frac{d x}{d y} ∣_{y = - 1} = e^{- 1 + 1} + C_{1} \overset{!}{=} 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 (x)

\Rightarrow f^{'} (x) = \frac{1}{x - x_{1} + 1}

Commented by York12 last updated on 15/May/24

Really, Really appreciate it

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