Question

e^x/y = 2x − y Solving by implicit

Answer 1

\[\Large\rm e^{x/y}=2x-y\]

Answer 2

Derivative huh? c:

Answer 3

Differentiating the exponential will be a little tricky.
Maybe we should start there.

Answer 4

You remember your e^x derivative, yes?

Answer 5

I believe so..

Answer 6

lol what is it? c:

Answer 7

(d/dx) e^x = e^x ? okay I dont remember :(

Answer 8

Ok good good.
Differentiating an exponential gives you the same thing back.

Answer 9

In our problem, since the "stuff" in the exponent position is more than just an x, we have to chain rule.

Answer 10

So here is how we start our problem, taking the derivative of each side with respect to x,
\[\Large\rm \color{royalblue}{\left(e^{x/y}\right)'}=\color{royalblue}{\left(2x-y\right)'}\]

Answer 11

so far I have, y.1-x.y/y^2=2-y ..after chain rule

Answer 12

Chaining.
\[\Large\rm e^{x/y}\color{royalblue}{\left(\frac{x}{y}\right)'}=\color{royalblue}{\left(2x-y\right)'}\]That's what you're getting for this chain here?

Answer 13

\[\Large\rm \left(\frac{x}{y}\right)'=\frac{x'y-xy'}{y^2}=\frac{y-xy'}{y^2}\]Does the second part of your numerator have a y'?
Maybe was a typo?
Check that again :)

Answer 14

Also with your right side,\[\Large\rm e^{x/y}\color{royalblue}{\left(\frac{x}{y}\right)'}=2-y'\]realize that anytime you take the derivative of `y with respect to x`, you're getting this y' or dy/dx popping up.

Answer 15

y.1-x.y'/y^2=2-y like this ? and ohhh okay

Answer 16

Any time you differentiate a `non-x` term, you'll get a prime showing up.
That's how I like to think about it at least :)

Ok good, don't forget your exponential though ^^

Answer 17

\[\Large\rm e^{x/y}\left(\frac{y-xy'}{y^2}\right)=2-y'\]

Answer 18

All of the differentiation is done at this point.
(which is actually, believe it or not, the easier part of this problem).

Now we have a bunch of nasty Algebra to get through in order to solve for y'.

Answer 19

Where you at girl? :O
You following along?

Any trouble getting to that point up there?

Answer 20

WAIT before you continue ! this thingy, y', we put that when y is with x, as you said before right?

Answer 21

Yes Im following along, Im writing in my note book as well so I remember :)

Answer 22

and where it says = 2x-y, we put y' so 2x-y',but why there as well ?

Answer 23

We put that prime thing whenever we take a derivative of a `non-x` value.

Example (differentiating both of these with respect to x):\[\Large\rm (x^2)'=2x\]\[\Large\rm (y^2)'\ne 2y\]We're differentiating y with respect to x, so we attach a y' to it.\[\Large\rm (y^2)'=2y~y'\]This is actually chain rule that causes this to happen.

If you think about the x^2 for a sec,\[\Large\rm (x^2)'=2x~x'=2x(1)\]The derivative of x with respect to x is just 1, so we don't pay any attention to it.
That step isn't necessary.

Answer 24

Oh alrighty ! I think Im starting to understand

Answer 25

so now we'll be solving for y' right?

Answer 26

Yes.

Answer 27

I would recommend multiplying both sides by y^2 as your first step.
Fractions are yucky.
It's nice if we can avoid them.

Answer 28

\[\Large\rm e^{x/y}(y-xy')=2y^2-y^2y'\]That step ok? :o

Answer 29

Yes :)

Answer 30

Mmmm what next? What do you think? >.<

Answer 31

sec, i gotta check on my pizza :OOO

Answer 32

LOOL okay ! Ill try to figure it out while you check on your pizza :P

Answer 33

Well we need to solve for y', so then hm, expnd the bracket, then solve for y'? :S

Answer 34

i see two y's

Answer 35

seems a bit hard

Answer 36

We have to try to get the y' s on one side

Answer 37

Expand the brackets, that seems like a good start.\[\Large\rm e^{x/y}y-x e^{x/y}y'=2y^2-y^2y'\]Yes, we have two y's.

Answer 38

Hmm let's uhhh, add the exponential term to the other side.
And we'll subtract 2y^2 from each side as well.

Answer 39

got it

Answer 40

Hmmmmm, what next? +_+

Answer 41

take y' common, then divide the first yet by xe^x/y-y^2 ?

Answer 42

Woops one of my y fell out of the problem, lemme fix that a sec D:\[\Large\rm y e^{x/y}-2y^2=x e^{x/y}y'-y^2y'\]Ok factor a y' out, then divide the bracket stuff to the other side? sounds goooood.

Answer 43

\[\Large\rm y e^{x/y}-2y^2=y'(x e^{x/y}-y^2)\]

Answer 44

\[\Large\rm y'=\frac{y e^{x/y}-2y^2}{x e^{x/y}-y^2}\]I'm gonna check it in Wolfram a sec, this is an easy one to make a careless mistake on.

Answer 45

Alrighty !

Answer 46

Oooo yay! We did it correctly! \c:/
Yayyy team!

Answer 47

So on a problem like this, if you get comfortable with the y' stuff, you'll notice that the differentiation becomes fairly easy.

You really need to be solid in your algebra skills to fully get through a problem like this!

Answer 48

Thank you so much ! I wish you were my math prof, you explain everything wayyyy better !

Answer 49

Hehe, aren't you sweet :3

Answer 50

Please do give your teacher a little bit of slack though :)
We spent 40 minutes on this single problem.

In class you're definitely more pressed for time, so it can be difficult squeezing in every little detail that you need.

Answer 51

That's true >.< But thanks again! :)