Derivative of implicit functions

Question

This problem got me confused. I'm supposed to find the value of the derivative of $$y\mathrm{e}^y=\mathrm{e}^{x+1}$$ at the point $x=0$.

I did find the derivate but I can't find its value because the derivate is $$\frac{y}{y+1}.$$ So basically there is no $x$ to plug in.

when $x=0$, what is $y$? And please use latex – daulomb Jan 13 '18 at 21:25 — daulomb, Jan 13 '18 at 21:25
It doesnt mention anything about y,just x. – E.Buzi Jan 13 '18 at 21:27 — E.Buzi, Jan 13 '18 at 21:27

score 0 · Answer 1 · answered Jan 13 '18 at 21:26

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we get $$y'e^{y}+ye^{y}y'=e^{x+1}$$ so $$y'=\frac{e^{x+1}}{e^y(y+1)}$$ for $$y+1\neq 0$$

answered Jan 13 '18 at 21:26

Dr. Sonnhard Graubner

95,283

Yes,i got to this point.what i wanted to know is how can i find the value of y' at the point x=0. – E.Buzi Jan 13 '18 at 21:28
we have $$y(0)e^{y(0)}=e$$ – Dr. Sonnhard Graubner Jan 13 '18 at 21:32
from here we get $$y(0)=1$$ – Dr. Sonnhard Graubner Jan 13 '18 at 21:33

score 0 · Answer 2 · answered Jan 13 '18 at 21:55

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You are ok. The left side of your equation is an increasing function of y and if x=0 the right side is 1 and the left side is 1 only at y=1. $\frac{y}{y+1}$ =$\frac{1}{2}$

answered Jan 13 '18 at 21:55

user439545

1,703
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Derivative of implicit functions

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