Find y'' at x=0 of:

xy + e^y = e.

Any and all help is greatly appreciated!

Question

Find y'' at x=0 of:

xy + e^y = e.

Any and all help is greatly appreciated!

anonymous · Answer

oops not clear, too many equal signs sorry

anonymous · Answer

$$y+xy'+e^yy'=0$$ that's better

anonymous · Answer

i get $$y'=-\frac{y}{x+e^y}$$ but it is late so check my algebra

anonymous · Answer

that's right. for y'

anonymous · Answer

whew

anonymous · Answer

then for $y''$ use the quotient rule

amonoconnor · Answer

Why does the "e" on the right side of the equals sign go to (0) rather than remaining as e? Wouldn't e^1 stay e^1 just as e^x stays e^x?

anonymous · Answer

$e$ is a number

anonymous · Answer

like $5$ or $\pi$

amonoconnor · Answer

Why is d/dx of e^x treated so specially then, in calculus?

anonymous · Answer

$e^x$ is  function, the exponential function but $e$ is just a number

anonymous · Answer

like the difference between $$f(x)=2^x$$ and $2$

anonymous · Answer

if $$f(x)=2^x$$ then $$f'(x)=2^x\ln(2)$$ whereas if $$f(x)=0$$ a constant then 
$$f'(x)=0$$

anonymous · Answer

ooops i meant if $$f(x)=2$$ then $$f'(x)=0$$

amonoconnor · Answer

Gotcha... I'm tired too I guess :p Thanks! I'm working through it again, with this "insight" lols. Give me just one sec.

amonoconnor · Answer

Is the derivative of the denominator (1-(e^y)*dy/dx) ?

zepdrix · Answer

derivative of the denominator?
I'm not sure what that means.
Do you mean the denominator of the derivative? :o

amonoconnor · Answer

derivative of the denominator of y', so g(x) I guess. The (x+e^y) term. Is it (1)?

amonoconnor · Answer

I'm using the quotient rule to find y'' from y'. :)

zepdrix · Answer

Oh oh oh :O

amonoconnor · Answer

I got y'' = (x+ e^y - y)/[(x+e^y)^2].
Do I realllyyy want to square that bottom, algebraically? Is that what I should do? :/

zepdrix · Answer

Derivative of the denominator: $\large\rm =(1+e^yy')$
Is that what you had? Addition, ya?

zepdrix · Answer

Yah your numerator is a lil botched up I think :)

amonoconnor · Answer

Damn it... X| Okay... starting over...

amonoconnor · Answer

This is why I said (1) at first, about d/dx(x + e^y)

zepdrix · Answer

I won't ruin any of the fun.
But I'll at least remind you of the "set up":$$\large\rm y'=\frac{y}{x+e^y}$$
$$\large\rm y''=\frac{\color{royalblue}{(y)'}(x+e^y)-y\color{royalblue}{(x+e^y)'}}{(x+e^y)^2}$$

zepdrix · Answer

For that online-calculator to accurately give you a derivative,
you probably need to specify that y is a function of x.

So you would input it like this: $\large\rm \frac{d}{dx}[x+e^{y(x)}]$
At least that's what you have to do on Wolfram Alpha

amonoconnor · Answer

And the derivative of "e^y" is, officially, (e^y)*y' ? Is that correct?

zepdrix · Answer

Mmmm yes, looks good.

amonoconnor · Answer

I'm getting:
$$y'' = \frac{-2y - \frac{x^2e^y}{x+e^y}}{(x+e^t)^2}$$

zepdrix · Answer

*I think* your y's should have canceled out on top.
Hmm lemme check my work again.

zepdrix · Answer

$$\large\rm y''=\frac{\color{royalblue}{(y)'}(x+e^y)-y\color{royalblue}{(x+e^y)'}}{(x+e^y)^2}$$
$$\large\rm y''=\frac{y'(x+e^y)-y(1+e^yy')}{(x+e^y)^2}$$
$$\large\rm y''=\frac{-y-y-ye^yy'}{(x+e^y)^2}$$Nope ^ I forgot the negative from y' :) hehe.
You were right there.

zepdrix · Answer

Plugging in for the other y',$$\large\rm y''=\frac{-2y-ye^y\frac{-y}{x+e^y}}{(x+e^y)^2}$$

zepdrix · Answer

Was that just a small typo in your second derivative maybe?
I see an x^2 in the second term, clearly it's a y^2 though, ya?

zepdrix · Answer

$$\large\rm y''=\frac{-2y+\frac{y^2 e^y}{x+e^y}}{(x+e^y)^2}$$

zepdrix · Answer

Oh..... they wanted you to evaluate this at a particular point?
Oh goodness, you're making this way hard on yourself then! :O
We don't need to do all this extra work.

amonoconnor · Answer

Yeah, whoops, that was a typo! :( 
And yep, at x=0...

zepdrix · Answer

Let's first find a corresponding y-coordinate for our x=0,$$\large\rm xy + e^y = e\qquad\to\qquad 0y+e^y=e$$So what's the y-coordinate? :)

amonoconnor · Answer

e-(e^y) ?

zepdrix · Answer

Hmm you're gettin too fancy.
So it simplifies to:$$\large\rm e^y=e^1$$

zepdrix · Answer

(Because the other thing was being multiplied by 0.)

zepdrix · Answer

The bases are the same, so the exponents have to be the same.
So $\large\rm y=1$, y a?

amonoconnor · Answer

Wait, so...$$xy + e^y = e$$
Can be evaluated to find "y" by doing: 
$$(0)y + (e) = (0)$$

amonoconnor · Answer

?

zepdrix · Answer

This: $\large\rm xy + e^y = e$
can be evaluated at x=0, $\large\rm 0y + e^y = e$
which will allow you to find a coordinate pair, (x,y).
Namely, the coordinate pair with x=0, (0,y).

amonoconnor · Answer

Ah... I see.

amonoconnor · Answer

So now I plug (0,1) into this?

amonoconnor · Answer

And that is y'', at the point, right?

zepdrix · Answer

Yes :) you can do that.
I was going to say that we could have avoided that really nasty process of quotient rule, and plugging in for the first derivative by using that (0,1) information earlier.

zepdrix · Answer

From here:$$\large\rm y'(x,y)=\frac{-y}{x+e^y}$$Evaluating the first derivative at (0,1) gives us,$$\large\rm y'(0,1)=\frac{-1}{0+e^1}=-e^{-1}$$That's your first derivative at (0,1).
Hold onto that information, and back up to this point,$$\large\rm y+y'(x+e^y)=0$$You COULD find your second derivative from here, it might be easier :)
You're already pretty deep into your problem with your method, so maybe it's not worth switching at this point.
But I show you just in case you're curious.

amonoconnor · Answer

Thank you:) That would be nicer in the future, for sure! So... I'm getting -3/e^2, but the book says 1/e^2.

zepdrix · Answer

$$\large\rm y'+y''(x+e^y)+y'(1+e^yy')=0$$Now DON'T go through the trouble of trying to solve for y'' just yet.
First plug in all the information.
x=0
y=1
y'=-e^-1$$\large\rm -e^{-1}+y''(0+e^1)-e^{-1}(1+e^1(-e^{-1}))=0$$Hmm the first derivative wasn't as easy to use as like... y'=5 or something :) lol
but it's still not too bad.

zepdrix · Answer

The book says 1/e^2?
Hmmm I'm getting -1/e^2...

amonoconnor · Answer

Me too... Ahhh! Ugg.

zepdrix · Answer

$$\large\rm y''(0,1)=\frac{-2(1)+\frac{(1)^2 e^1}{0+e^1}}{(0+e^1)^2}=\frac{-2+1}{e^2}$$Hmm :d

zepdrix · Answer

Imma check the other way that I did it.

amonoconnor · Answer

Just checked my math and signs again, and am getting -1/x^2 ... :(

amonoconnor · Answer

*Sorry: -1/e^2

zepdrix · Answer

Ooo I'm getting 1/e^2 from my other approach :o oh boy

amonoconnor · Answer

oh god.. did I do it wrong again? :/

freckles · Answer

hey

freckles · Answer

the mistake is just leaving off the - in the y'

amonoconnor · Answer

I'm so frustrated with having these little issues where I can get like, 75% of the way through a problem but then can't get the damn answer! It' not like I don't get the foundations, chain rule, product rule, etc., there're these little situational questions where I just ostensibly don't see how pellet works. God.

freckles · Answer

$$y'=-\frac{y}{x+e^y}  $$
somehow that - got dropped

freckles · Answer

just bring down the - to y''

amonoconnor · Answer

Okay, will do!

amonoconnor · Answer

Can you guys help me with another one?

zepdrix · Answer

$$\large\rm y''=\frac{\color{royalblue}{(-y)'}(x+e^y)-(-y)\color{royalblue}{(x+e^y)'}}{(x+e^y)^2}$$Yah I missed the negatives in my second derivative setup :( woops.

zepdrix · Answer

Sure :)

zepdrix · Answer

We won't make the same mistakes this time :D

freckles · Answer

i do agree it is hard being perfect

freckles · Answer

i was doing some ugly math stuff the other day and it involved adding 2 and 2
and for some reason I put 3 instead of 4 which made my end result a little off

freckles · Answer

i know that sounds dumb

amonoconnor · Answer

I know exactly how you feel... Ugg.

zepdrix · Answer

XD