Implicit Differentiation

[slim pier]

I don't know why we do this nor do I know how to do it. Like wouldn't this use just the product rule as the deriviative of x(e^y) just x'(e^y)+(x)e^y'

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[rapid jolt]

it looks like to me, the best approach would be logarithmic differentiation.

y = xe^y
ln(y) = ln(xe^y)
ln(y) = ln(x) + ln(e^y)
ln(y) = ln(x) + y

then you take the derivative of both sides of the equation from here and apply chain rule

[neat salmon]

Actually, you've got it right there. Now you've just got an equation to solve.

[rapid jolt]

this works even if y or x is negative due to how quotient and product rule work, some little details tho

[slim pier]

no this isn't my work technie

this is teachers

[neat salmon]

Okay, so then what's the derivative of x?

With respect to x?

[slim pier]

so when we do with the respect to x do we just add x' to the x value

[neat salmon]

...what?

[slim pier]

i don't know how to do differention to the respect to a value

[neat salmon]

"With respect to x" just means we're treating x as the variable, and everything else is either a constant or a function of x.

[slim pier]

okay so the new equation would just be x=y(e^x)

so x is the function to respect of y

[neat salmon]

...what?

No.

[slim pier]

other way

[neat salmon]

What is the derivative of x?

[slim pier]

1

[neat salmon]

Right. That's the derivative of x with respect to x.

[slim pier]

oh so were only taking the derivate of x

[neat salmon]

No, we're taking the derivative of everything.

Look, do you understand what a derivative is, like, conceptually?

[slim pier]

NO

oops

no

[neat salmon]

Okay.

So imagine we have a curve.

[slim pier]

yea

[neat salmon]

We want to know how fast the curve is changing at a specific point.

[slim pier]

so instanous velocity/change right

[neat salmon]

...you said you didn't understand.

[slim pier]

yea i memorized it but it doesn't make sense

[neat salmon]

Okay, then just listen.

[slim pier]

okay!

[neat salmon]

So we want to know how fast the curve is changing at a given point.

The problem is, the curve doesn't "change" "at a point".

Just like the phrase "instantaneous velocity" doesn't make sense, because velocity is a rate of change of position, and position doesn't change instantly.

[slim pier]

exactly!

[neat salmon]

So we want to find the slope of a curve at a point, but we can't, because, well, we only have one point.

We can calculate the slope between two points, though.

And if both of those points are on our curve, then we call that a secant line.

[slim pier]

yes

which is the average line

[neat salmon]

Right.

So ideally, this secant line would approximate the tangent line.

And we can make it do that, by moving the second point closer to the first.

[slim pier]

oh

as the secent line gets closer to a point it becomes the tangent line

oops

as the two points gets closer to each other it becomes tangent line

[neat salmon]

It better approximates the tangent line.

[slim pier]

okay

[neat salmon]

What we would really like is some kind of precise, well-defined, rigorous mathematical method to get as close as we like to the tangent line. Arbitrarily close.

This is the concept we call a limit.

[slim pier]

okay

[neat salmon]

So were you taught the limit definition of a derivative? $f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$?

[sudden geode]

Techie Literate

[slim pier]

yes and no, i was thought the limit definition

which is as the x value approaches a certain number what is the y value becoming

[neat salmon]

$\lim_{x \to a}\frac{f(x) - f(a)}{x - a}$?

[sudden geode]

Techie Literate

[slim pier]

this is the same

[neat salmon]

...as what you were taught, or as what I said before?

[slim pier]

this equation is using two points

while the other equation is using the whole equation

[neat salmon]

Okay, I think you're trying to tell me that you were taught the definition of a limit.

[slim pier]

yes

[neat salmon]

Which, you kind of weren't.

The real definition is more rigorous than what you were taught, but it's not strictly necessary for you to know.

[neat salmon]

I mean, that would be equal to f'(a).

[slim pier]

okay

[neat salmon]

So the intuition here in this limit is that the point we care about is (x, f(x)), and h is the distance between that point and the second point we're using to approximate the slope.

[slim pier]

okay

[neat salmon]

But this limit is time consuming to compute directly every time, so we just use it to prove the derivative rules and use those instead.

[slim pier]

mm okay

[neat salmon]

Now, circling back around to your problem, the issue is that you don't actually have y as a function of x.

This equation does still describe a curve, but it's just not in a nice form that's easy to manipulate.

[slim pier]

we do have a function y that respects x

[neat salmon]

Except we have y on both sides.

[slim pier]

oh okay

[neat salmon]

We could technically isolate it, but that would require a non-elementary function that would just be more confusing for you.

[slim pier]

okay

[neat salmon]

But here's the thing.

[slim pier]

Wait, rq

[neat salmon]

Okay?

[slim pier]

so if we have this same equation but instead of of having e^y if we had e^z could we just do the product rule to find the derivative of y

[neat salmon]

Like, z would be a third variable?

Or would z be a constant? Because then e^z is just a constant coefficient and we're looking at a line.

[slim pier]

oops

i mean e^x

[neat salmon]

Oh, yeah.

[slim pier]

so we would just do the product rule

but since y isn't a function as there is y in both sides then we have to do implicit differation

[neat salmon]

Well, y is a function, it's just not represented nicely.

And it may not be a function per se.

[slim pier]

and thats why we have to do implicir diff

[neat salmon]

Right.

[slim pier]

okay contiue

[neat salmon]

See, here's how equations that describe curves work; they have x and y in them, and then you plot all the points (a, b) such that when x = a and y = b, the equation is true.

So this equation does describe a curve, it just doesn't do so in standard y = f(x) form. Importantly, however, the slope of the tangent line will still be y'.

[slim pier]

okay

[neat salmon]

So the rule is basically that we treat y as a function of x, but we don't know what function of x.

[slim pier]

k

[neat salmon]

So you were correct that the very first step is to use the product rule, (xe^y)' = (x)' * e^y + x * (e^y)'.

[slim pier]

okay

[neat salmon]

And that would be equal to y'.

So now what we want is to solve for y', in terms of x and y.

So what does our equation look like?

[slim pier]

my closes thought is point slope form

[neat salmon]

No, starting from the equation given in the problem, differentiate both sides with respect to x.

That is, treat x as the variable and y as a function of x.

[slim pier]

if we differentaite both sides we get

1=2e^y

[neat salmon]

...show your work.

[slim pier]

i mean e^yx +e^y

give me a second im about to show work

[neat salmon]

Maybe some different notation will help.

We want to take d/dx on both sides.

So what's d/dx (y)?

[slim pier]

what does that even mean

d/dx means the derivative y to respect x

and why would we even do thatg

[neat salmon]

No, dy/dx means that.

[slim pier]

oh

so d/dx is the derivative to respect x

[neat salmon]

d/dx (y) means take the derivative of y with respect to x, dy/dx is the derivative of y with respect to x.

[slim pier]

isn't that the same thing!?

[neat salmon]

No. The first is an operation, the second is the output of the operation.

[slim pier]

okay

[neat salmon]

If I hand you a deck of cards and tell you to deal a five card hand, my instruction to deal five cards is not equivalent to the actual five cards that end up on the table.

[slim pier]

okay

[neat salmon]

Or better yet, just rewrite the equation with y(x) instead of y.

[slim pier]

Oh i understand

so d/dx is like a minus sign in which you can use for a specific number/variable

and dy/dx is the derivative of the whole equation

so what do i do after the 1=e^y(x)+e^y

[neat salmon]

Nothing because you made a mistake. Go back to the previous step and rewrite y as y(x) to make it explicit that y is a function of x and see if that shakes your intuition loose.

[slim pier]

ohhh

so i do the product rule

and we get x+y=e^y(x)+e^y

now what

?

[neat salmon]

...no.

Not y times x. y of x.

[slim pier]

whats the difference

like

[neat salmon]

What's the difference between f times x and f of x?

[slim pier]

u told me to replace to y to y(x)

to make me think of something

and i did

and the only thing i could think of is product rule

[neat salmon]

...okay. y = f(x).

[slim pier]

i agree

[neat salmon]

So replace y with f(x).

[neat salmon]

On both sides.

[slim pier]

okay i did

[neat salmon]

$f(x) = xe^{f(x)}$

[sudden geode]

Techie Literate

[slim pier]

what my instructor did was y'= (xe^y)' (y')

[neat salmon]

No, he didn't.

[slim pier]

the top row was the producr rule

[neat salmon]

Do you see how this is not what you said?

[slim pier]

times y'

[neat salmon]

No, the entire thing isn't times y', you multiply the second term by y' because of the chain rule, because y is a function of x.

[slim pier]

oh okay

[neat salmon]

That's the whole thing I was trying to get you to understand, that y' is not 1.

[slim pier]

chain rule: (3a+2b)^7'= 7(3a+2b)^6 * (5)

[neat salmon]

...what?

[slim pier]

wait im just trying to see how u use the chain rule for

give me one second

oh OKAY

Im back

so why do we use the chain rule in this scenarion

why don't we just do the product rule

[neat salmon]

Because. Y. Is. A. Function. Of. X.

[slim pier]

it's because theres a y in both sides right?

im sorry

[neat salmon]

No. It's because y is a function of x.

[slim pier]

okay

i think i got it

thanks for your help and sorry for the stress

[neat salmon]

Like, okay.

[slate kindle]

I can teach you an informal way of solving this if you want

[neat salmon]

Just to solidify your intuition, so that you don't go around just assuming this by rote.

If we look at our original equation, and we pick a value for x, then that restricts y, right?

[slim pier]

yes

[neat salmon]

So y depends on x.

[slim pier]

yes

[neat salmon]

And this dependence is the reason we treat y as a function of x when we differentiate.

Which is why the derivative of y with respect to x is y', not 1 or anything else.

[slim pier]

okay

[neat salmon]

Do you understand?

[slim pier]

yes

[neat salmon]

Okay, good, I'm glad.

[slim pier]

thanks

[neat salmon]

I didn't want you to just agree with me because you thought I was getting angry.

I wanted you to understand the actual line of thought being followed.

[slim pier]

+clos

+close