
English: 
So let's take another
implicit derivative
of the somewhat
crazy relationship.
And I've graphed the
relationship here.
As you can see, it is
actually quite bizarre.
e to the x times y squared
is equal to x minus y.
This is some at least in the
range that's depicted here,
the x's and the y's that
satisfy this relationship.
Well let's take the
derivative of both sides.
So we'll apply our derivative
operator to both sides.
And actually this
is a good chance
to maybe explore some
different notation.
We tend to be using
that notation,
but oftentimes you'll see
a derivative operator that
just looks like a big
capital D. So maybe we'll
do that right over here.
So let me make it clear.
This is the equivalent
of saying d over dx.
I'll just use this
big capital D operator
so that you are familiar
with the notation.
And instead of using dy
dx for the derivative of y

Thai: 
 
ลองหาอนุพันธ์โดยนัยอีกตัว
ของความสัมพันธ์เพี้ยนๆ นี้ดู
และผมวาดกราฟความสัมพันธ์ตรงนี้
อย่างที่คุณเห็น มันแปลกจริงๆ
e กำลัง x คูณ y กำลังสองเท่ากับ x ลบ y
นี่คือช่วงที่วาดมา
x กับ y ที่เป็นไปตามความสัมพันธ์นี้
ลองหาอนุพันธ์ทั้งสองด้านกัน
เราจะใช้ตัวดำเนินการอนุพันธ์ทั้งสองข้าง
และนี่เป็นโอกาสอันดี
เพื่อสำรวจสัญลักษณ์อื่นบ้าง
เรามักใช้สัญลักษณ์นั้น
แต่หลายครั้ง คุณจะเห็นตัวดำเนินการอนุพันธ์ที่
ดูเหมือนตัว D ใหญ่ บางทีเราจะ
ใช้มันตรงนี้
ขอผมบอกให้ชัดนะ
นี่เทียบเท่ากับการบอกว่า d ส่วน dx
ผมจะใช้แค่ D ใหญ่เป็นดำเนินการ
คุณจะได้คุ้นเคยกับสัญลักษณ์
และแทนที่จะใช้ dy/dx แทนอนุพันธ์ของ y

Bulgarian: 
Нека да решим още един пример 
за неявно диференциране
на донякъде щура зависимост.
Направил съм графиката 
на тази зависимост ето тук.
Както може да видиш, 
всъщност е доста странна.
e на степен x по y е равно на x – y.
Това е част от нея, поне в интервала, 
който е изобразен,
за x и y стойностите, които 
удовлетворяват  даденото уравнение.
Нека да намерим производната 
на двете страни на уравнението.
Записвам означението за производна  d/dx в двете страни на уравнението.
Това всъщност е добра възможност
да изследваш 
някакво различно означение.
Имаме склонност да използваме
ето това означение,
но често ще виждаш означение 
за производна,
което изглежда като главно D (диференциален оператор). 
Така че, може би
ще го използваме точно тук.
Нека да го изясня.
Това е еквивалентно на 
означението d/dx.
Просто ще използвам 
това означение с главно D,
така че да се запознаеш 
с означението.
Вместо да използвам dy/dx 
за производната на y

Czech: 
Zkusme implicitně zderivovat
tento poněkud šílený vztah.
Tady je graf, na kterém můžete vidět,
že jde skutečně o docela bizarní vztah.
e na (x krát y na druhou)
se rovná x minus y.
Tady máme některá x a y, která se vešla
do grafu a vyhovují tomuto vztahu.
Pojďme obě
strany zderivovat.
Na obě strany tedy
použijeme operátor derivace.
Tohle je možná příhodná chvíle k tomu,
abychom si ukázali různá značení.
Většinou používáme
tohle značení,
ale často se můžete setkat s operátorem
derivace zapsaným jako velké tiskací D,
takže to teď
zapišme třeba takto.
Raději to
ujasním.
Toto je totéž jako
napsat d lomeno dx.
Použiji zde operátor zapsaný
jako velké tiskací D,
abyste si zvykli
i na tento zápis.

Korean: 
 
이상한 형태의 음함수의
도함수를 구해봅시다
여기 이 식의 그래프를 그렸습니다
굉장히 이상해보입니다
e^(xy²)＝x－y
그래도 이것을 보면
이 식을 만족하는 x 와 y 가
이 범위 안에는 있습니다
양변을 미분해봅시다
양변을 미분해봅시다
새로운
표기법을 알아봅시다
이렇게 쓰는것이 일반적이지만
가끔
D 를 볼 수 있을 것입니다
여기도 그렇게 하겠습니다
다시 한 번 말하자면
이것은 d/dx 와 같은 표현입니다
D 에 익숙해지기 위해서
D 를 이용하겠습니다
그리고 y의 도함수를 표현할 때

Portuguese: 
Vamos tomar a derivada
de uma função implícita,
que é de certa forma
uma função meio doida.
Eu tracei a função aqui.
Dá para notar que ela
é bem bizarra.
e elevado a x vezes o quadrado de y
é igual a x menos y.
Pelo que está representado nesse
gráfico, esses são
os valores x e y que satisfazem
essa relação.
Bem, vamos tomar as derivadas
de ambos os lados.
Aplicando o operador derivada 
nos dois lados.
Essa é uma boa 
oportunidade
para explorarmos uma
notação diferente.
Costumamos usar 
essa notação,
mas às vezes você verá
o operador de derivada
representado por
um D maiúsculo.
Então vou usá-lo bem aqui.
Deixando bem claro.
Isso equivale 
a d sobre dx.
Vou usar somente o 
D maiúsculo,
para você se familiarizar
com essa notação.
E ao invés de usar dy sobre dx
para representar a derivada de y

Korean: 
dy/dx 대신 y'이라고 쓰겠습니다
이 새로운 표기법으로
연습해볼 필요가 있습니다
여기 있는 것의 도함수를 구해봅시다
합성함수의 미분법을
이용해야 합니다
사실 합성함수의 미분법을
여러번 사용해야 합니다
e의 거듭제곱 꼴의 도함수는
그 원래 식에
지수를 x에 대해 미분한 것을
곱해주면 됩니다
여기에 xy² 의 도함수를 곱한 것이
좌변이 됩니다
아직 끝나지 않았습니다
우변을 보면 x의 도함수는 1이 되고
y를 x에 대해 미분한 것은
dy/dx 인데 이것 대신에
y' 이라고 적겠습니다
저는 이렇게 쓰는 것을 더

Portuguese: 
com relação a x, vou
escrever somente y linha.
Assim praticamos
com uma notação diferente.
Vamos tomar a derivada 
disso bem aqui.
Aplicaremos a regra da cadeia.
Na verdade, vamos usar
a regra de cadeia
várias vezes aqui.
A derivada de e elevado
a alguma coisa com relação
à essa alguma coisa 
será e elevado
a essa coisa
vezes a derivada
dessa coisa 
com relação a x.
Então, vezes a derivada
de xy ao quadrado.
Esse é o lado esquerdo.
Não terminamos com a
derivada ainda.
E no lado direito a
derivada de x é igual a 1.
E a derivada de y
com relacão a x
será igual a menos - ou podemos
escrever - dy sobre dx negativo.
Mas ao invés de escrever dy sobre dx,
vou escrever y linha.
Você pode ver que eu gosto dessa
notação e dessa notação

Bulgarian: 
спрямо x, просто 
ще го записвам като y'.
За да се упражним малко
с различно означение.
Нека да намерим производната 
на този член ето тук.
Ще приложим верижното правило.
Всъщност ще приложим 
верижното правило
множество пъти в този пример.
Производната на e на степен нещо
спрямо това нещо, ще бъде
e на степен същото нещо, 
и умножено по производната
на това нещо спрямо x.
Следователно по производната 
на x по y на квадрат.
Това е нашата лява страна.
Не сме приключили все още с 
търсенето на производни.
А от нашата дясна страна, 
производната на x е равна просто на 1.
И производната спрямо x на y
ще бъде просто минус...
мога да запиша... минус dy/dx.
Но вместо да записвам dy/dx, 
ще запиша y'.
Както можеш да видиш, 
харесвам това и това означение

Thai: 
เทียบกับ x ผมจะเขียนมันว่า y ไพรม์
เราจะได้ฝึก
ใช้สัญลักษณ์ต่างๆ กัน
ลองหาอนุพันธ์ของสิ่งนี้ตรงนี้
เราจะใช้กฎลูกโซ่
ที่จริง เราจะใช้กฎลูกโซ่
หลายครั้งตรงนี้
อนุพันธ์ของ e กำลังอะไรสักอย่างเทียบกับ
อะไรสักอย่างนั้น จะเท่ากับ e
กำลังอะไรสักอย่างนั้นคูณอนุพันธ์
ของอะไรสักอย่างนั้นเทียบกับ x
คูณอนุพันธ์ของ xy กำลังสอง
นั่นคือด้านซ้ายมือ
เรายังหาอนุพันธ์ไม่เสร็จ
ทางขวามือ อนุพันธ์ของ x ก็แค่ 1
และอนุพันธ์เทียบกับ x ของ y
จะเท่ากับลบ -- ผมเขียนว่า -- ลบ dy/dx ได้
แต่แทนที่จะเขียน dy/dx ผมเขียน y ไพรม์ได้
อย่างที่บอก ผมชอบสัญลักษณ์นี้ และสัญลักษณ์นี้

Czech: 
Namísto dy lomeno dx, když
chceme napsat derivaci y podle x,
budu psát y s čárkou,
abychom si trochu
procvičili různá značení.
Spočítejme teď
derivaci tohohle.
Použijeme vzorec pro derivaci složené
funkce, dokonce ho použijeme několikrát.
Derivace (e na něco) podle
toho něčeho je e na to něco,
což ještě musíme vynásobit
derivací toho něčeho podle x,
tedy krát derivace z
(x krát y na druhou).
Tak vypadá
levá strana,
ale ještě jsme
nespočítali tuhle derivaci.
Na pravé straně...
Derivace x je 1
a derivace podle ‚x‘ z ‚y‘ se rovná
minus dy lomeno dx,
ale místo dy lomeno dx
napíšu y s čárkou.
Jak asi vidíte, mám
raději tenhle zápis,

English: 
with respect to x, I'm just
going to write that as y prime.
So we get a little
bit of practice
with different notation.
So let's take the derivative
of this thing right over here.
Well we're going to
apply the chain rule.
Actually, we're going
to apply the chain rule
multiple times here.
The derivative of e to
the something with respect
to that something
is going to be e
to the something
times the derivative
of that something
with respect to x.
So times the derivative
of xy squared.
So that's our left-hand side.
We aren't done taking
the derivative yet.
And on our right-hand side,
the derivative of x is just 1.
And the derivative
with respect to x of y
is just going to be minus-- or
I could write-- negative dy dx.
But instead of writing dy dx,
I'm going to write y prime.
As you can tell, I like this
notation and this notation

Bulgarian: 
повече защото става ясно,
че търся производната спрямо x.
Тук просто трябва да предположим,
че търсим производната спрямо x.
Тук трябва да предположим, че 
това е производната на y
спрямо x.
Без значение, нека в този пример 
да се придържаме към това означение тук.
Всъщност, нека да направя всички y',
т.е. всички производни на y спрямо x...
Нека да ги направя в розово, 
така че да мога да ги следя.
Още веднъж, това ще бъде
равно на e на степен x по y^2, умножено по 
производната на този член.
За производната на този член 
може просто
да използваме правилото за намиране на
производна на произведение и верижното правило
ето тук.
Производната на x е просто 1,
умножено по втората функция.
Ще бъде умножено по y на квадрат.
След това ще направим следното. 
Ще добавим
произведението на първата функция,
която е това x по производната 
на y на квадрат спрямо x.

Portuguese: 
mais ainda, porque deixa 
mais explícito
que eu estou tomando a
derivada com relação a x,
Aqui teremos que assumir que
a derivada é em relação a x.
Aqui teremos que assumir que
essa é a derivada de y
com relação a x.
Enfim, vamos continuar usando
essa notação bem aqui.
Vou fazer todos os meus y linha,
todas as minhas derivadas de y
com relação a x,
na cor rosa, para 
uniformizar todos eles.
Mais uma vez, isso será
igual a e elevado a xy ao quadrado
vezes a derivada disso aqui.
Para essa derivada,
podemos simplesmente usar,
o produto e um pouco de 
regra de cadeia aqui.
Então a derivada de x é igual a 1
multiplicado pela segunda função.
Isto é, multiplicado por
y ao quadrado.
E a isso adicionamos
o produto da primeira função,
que é x vezes a derivada de 
y ao quadrado, com relação a x.

Korean: 
선호하는데
이 방법이 더 분명해보입니다
x에 대해 미분하고 있다는 것을
알고 있어야 합니다
여기는 x에 대한 y의 도함수임을
알아야 합니다
우선 이것에 집중을 해보겠습니다
여기 있는 모든 y' 과
y의 x에 대한 도함수를
핑크색으로 보기 쉽게
하겠습니다
다시 말하자면 이것은
e^(xy²)와 지수의 도함수를
곱한 것이 됩니다
이것의 도함수를 구하려면
합성함수의 미분법과 곱의 미분법이
필요합니다
합성함수의 미분법과 곱의 미분법이
필요합니다
x를 미분하면 1이 되므로
여기 y² 이 됩니다
그 다음에는
첫 번째 함수
x와 y²의 x에 대한 도함수를
곱해야 합니다

Thai: 
มากกว่าเพราะผมบอกได้ตรงๆ
ว่าผมกำลังหาอนุพันธ์เทียบกับ x
ตรงนี้ เราต้องสมมุติว่าเรากำลัง
หาอนุพันธ์เทียบกับ x
ตรงนี้ เราต้องสมมุติว่ามันคืออนุพันธ์ของ y
เทียบกับ x
แต่ช่างเถอะ ลองใช้สัญลักษณ์นี่ตรงนี้ต่อ
ที่จริง ขอผมเขียน y ไพรม์ทุกตัว
อนุพันธ์ของ y เทียบกับ x ทุกตัว
ขอผมใช้สีชมพู ผมจะได้ติดตามมันได้
เหมือนเดิม อันนี้จะ
เท่ากับ xy กำลังสอง คูณอนุพันธ์ของตัวนี้
อนุพันธ์ของตัวนี้ เราแค่
ใช้กฎผลคูณ กับกฎลูกโซ่นิดหน่อย
ตรงนี้
อนุพันธ์ของ x ก็แค่ 1 คูณฟังก์ชันที่สอง
มันจะคูณ y กำลังสอง
แล้วจากนั้น เราจะบวก
ผลคูณของฟังก์ชันแรกซึ่ง
ก็คือ x นี้คูณอนุพันธ์ของ 
y กำลังสองเทียบกับ x

Czech: 
protože je z něho jasné,
že derivujeme podle x,
zatímco zde musíme předpokládat,
že derivujeme podle x,
stejně tak zde musíme předpokládat,
že jde o derivaci y podle x.
Držme se ale
tohoto značení.
A pokaždé...
Všechna y s čárkou, tedy všechny derivace
podle x, nebo spíše derivace y podle x,
napíšu růžovou, abych
lépe viděl, kde jsou.
Toto se rovná e na (x krát y na druhou),
to celé krát derivace tohohle.
Na derivaci tohohle použijeme vzorce
pro derivaci součinu a složené funkce.
Derivace x je 1, což musíme vynásobit
druhou funkcí, tedy krát y na druhou,
a k tomu musíme přičíst
součin první funkce, což je x,
krát derivace
(y na druhou) podle x.

English: 
more because it
makes it explicit
that I'm taking the
derivative with respect to x.
Here, we just have
to assume that we're
taking the derivative
with respect to x.
Here, we have to assume
that's the derivative of y
with respect to x.
But anyway let's stick with
this notation right over here.
Actually, let me make
all of my y primes,
all my derivatives of
y with respect to x,
let me make them pink
so I keep track of them.
So once again,
this is going to be
equal to e to the xy squared
times the derivative of this.
Well the derivative
of this, we can just
use the product and actually
a little bit of the chain rule
here.
So the derivative of x is just
1 times the second function.
So it's going to
be times y squared.
And then to that,
we're going to add
the product of the
first function which
is this x times the derivative
of y squared with respect to x.

Thai: 
นั่นจะเท่ากับอนุพันธ์ของ y กำลังสอง
เทียบกับ y ซึ่งก็แค่ 2y
คูณอนุพันธ์ของ y เทียบกับ
x ซึ่งเราเขียนได้ตอนนี้เป็น y ไพรม์
 
แล้วมันจะเท่ากับ 1 ลบ y ไพรม์
 
และอย่างที่เราทำมา ตอนนี้เรา
ต้องแก้หา y ไพรม์
ลองแจกแจงพจน์ยกกำลังนี้ e
กำลัง xy กำลังสองนี้
แล้วเราได้ e บางทีผมควรบอกว่า 
y กำลังสองคูณ e
กำลัง xy กำลังสอง
แค่นั่นแหละ
บวก 2xy e กำลัง xy กำลังสอง
y ไพรม์, อนุพันธ์ของ y เทียบกับ x

English: 
Well that's going to be
the derivative of y squared
with respect to y, which
is just going to be 2y
times the derivative
of y with respect
to x, which we are now
writing as y prime.
And then that's going to be
equal to 1 minus y prime.
And like we've
been doing, we now
have to just solve for y prime.
So let's distribute
this exponential, this e
to the xy squared.
And we get e, or maybe I
should say y squared times e
to the xy squared.
So that's that.
Plus 2xye to the xy squared.
y prime, the derivative
of y with respect to x,

Portuguese: 
Isso é igual à derivada de 
y ao quadrado com
relação a y, o que
dá 2 vezes y
vezes a derivada de y 
com relação a x,
o que estamos escrevendo
como y linha.
Isso será igual a 1 menos y linha.
E como vínhamos fazendo 
até agora,
resolvemos isso para y linha.
Então vamos distribuir essa
exponencial, esse e
elevado a x vezes o quadrado de y.
E teremos e, ou melhor dizendo,
y ao quadrado vezes e,
elevado a x vezes o quadrado y.
Isso está feito.
Mais 2 xye elevado a 
x y ao quadrado vezes
y linha, a derivada de y
com relação a x,

Korean: 
y²을
y에 대해 미분하면 2y 이고
y를 x에 대해 미분한 것을
곱해야 하는데
그것이 y' 이 됩니다
 
우변은 1－y' 입니다
 
이제 y' 에 대해서
정리해주면 됩니다
e^(xy²) 을
분배시켜 봅시다
그러면 e^(xy²)y² 이 됩니다
그러면 e^(xy²)y² 이 됩니다
그러면 y² e^(xy²)이 됩니다
그 다음 항은 2xye^(xy²)y' 가 됩니다
그 다음 항은 2xye^(xy²)y' 가 됩니다

Bulgarian: 
Това ще бъде производната на y^2
спрямо y, което ще бъде просто 2y,
по производната на y спрямо x,
което сега записваме като y'.
Приравняваме на дясната страна, 
която ще е равна на 1 – y'.
Това ще е равно на 1 – y'.
И както досега сме решавали 
уравненията,
сега просто ще намерим от тук y'.
Нека да разкрием скобите и умножим
по този израз със степен, т.е. това e
на степен x по y на квадрат.
И вземаме 'e', или следва да кажа y^2
по е на степен x по y^2.
Става дума за това.
Плюс 2xy по е на степен xy^2.
y', производната на y спрямо x,

Czech: 
Ta se rovná derivaci (y na druhou)
podle y, což je 2 krát y,
vynásobené
derivací y podle x,
kterou nyní značíme
jako y s čárkou.
Tohle celé je rovno
1 minus y s čárkou.
Jako jsme to dělali už dříve, musíme
nyní osamostatnit y s čárkou.
Roznásobme nejdřív touhle exponenciální
funkcí, tímhle e na (x krát y na druhou),
čímž dostaneme, že e...
Nebo bych spíš měl říct, že (y na druhou)
krát e umocněné na (x krát y na druhou),
to je tenhle člen,
plus 2 krát x krát y krát e umocněné na
(x krát y na druhou) krát y s čárkou,
což je derivace y podle x.

Thai: 
เท่ากับ 1 ลบอนุพันธ์ของ y เทียบ
กับ x
ตอนนี้ลองนำ y ไพรม์ทุกตัวไว้ข้างหนึ่ง
ลองบวก y ไพรม์ทั้งสองข้าง
ลองบวก -- ขอบอกให้ชัด
ผมกำลังบวก 1 y ไพรม์ทั้งสองข้าง
บวก y ไพรม์ทั้งสองข้างกัน
แล้วลองลบตัวนี้จากทั้งสองด้าน
ลองลบ y กำลังสอง e กำลัง xy กำลังสอง
ลบจากทั้งสองด้าน
เราจะลบ y กำลังสอง e กำลัง xy กำลังสอง
แล้วเราเหลือ 2xy e กำลัง xy
กำลังสอง บวก 1 คูณ y ไพรม์
เราได้ y ไพรม์เท่านี้ แล้ว
เราบวกอีก 1 y ไพรม์ 
เราจึงได้เท่านี้บวก 1 y ไพรม์

English: 
is equal to 1 minus the
derivative of y with respect
to x.
Now let's get all of our
y primes on one side.
So let's add y
prime to both sides.
So let's add-- and
just to be clear,
I'm adding one y
prime to both sides.
So let's add a y
prime to both sides.
And let's subtract this
business from both sides.
So let's subtract y
squared e to the xy squared
subtracting from both sides.
So we're going to subtract y
squared e to the xy squared.
And we are left
with 2xye to the xy
squared plus 1 times y prime.
We had this many
y primes and then
we add another 1y prime so we
have this many plus 1 y primes.

Korean: 
이것이 1－y' 과 같습니다
이것이 1－y' 과 같습니다
모든 y' 을 한쪽으로 몰아봅시다
양변에 y' 을 더하면 될 것 같습니다
양변에
y'을 하나씩만 더해주면 됩니다
y'을 하나씩만 더해주면 됩니다
그리고 양변에서 이것을 뺍시다
양변에서 y² e^(xy²)를 빼주면 됩니다
양변에서 y² e^(xy²)를 빼주면 됩니다
양변에서 y² e^(xy²)를 빼주면 됩니다
남은 것은 (2xye^(xy²)＋1)y' 입니다
남은 것은 (2xye^(xy²)＋1)y' 입니다
이만큼의 y'이 있다가
한개를 더해줬으므로
원래의 양보다 하나 더 많아졌습니다

Portuguese: 
é igual a 1 menos 
a derivada de y com
relação a x.
Vamos passar todos os nossos
y linha para um só lado.
Adicionando y linha dos dois lados.
Vamos adicionar, e para
ficar bem claro,
estou adicionando um y linha 
nos dois lados.
Somando um y linha de cada lado.
E subtraindo isso aqui dos dois lados.
Vamos subtrair y ao quadrado
vezes e elevado a xy ao quadrado,
subtraindo dos dois lados.
Vamos subtrair o y ao quadrado
vezes e elevado a x y ao quadrado.
E ficamos com 2xye elevado 
a x y ao quadrado
mais 1 vezes y linha.
Tínhamos todos esses y linha e
agora somamos 1y linha, então
temos todos esses mais 1 y linha.

Bulgarian: 
е равна на 1 минус производната на y
спрямо x.
Сега нека да прехвърлим 
всички членове y' от едната страна.
Нека да прибавим y' към двете страни.
Нека да го направим. И, просто за яснота,
добавям 1 по y' към двете страни на уравнението.
Нека да прибавим y' към двете страни.
Нека да извадим този член 
от двете страни.
И нека да извадим y^2 по e на степен xy^2
от двете страни на уравнението.
Ще извадим y^2 по 
e на степен xy^2.
И остава 2xy по е на степен x
по y^2 плюс 1, цялото по y'.
Имахме ето толкова y' и тогава
добавихме друго 1 по y', така че имаме 
всички тези членове плюс 1 по y'.

Czech: 
To se rovná 1 minus
derivace y podle x.
Nyní přesuneme všechna
y s čárkou na jednu stranu,
takže k oběma stranám
přičteme y s čárkou.
Přičteme...
Aby to bylo jasné, tak k oběma
stranám přičítám jedno y s čárkou.
K oběma stranám
přičteme y s čárkou
a od obou stran
odečteme tento výraz.
Od obou stran tedy odečteme (y na druhou)
krát e umocněné na (x krát y na druhou).
Odečítáme od obou stran,
takže i zde odečteme (y na druhou)
krát e umocněné na (x krát y na druhou).
Zbyde nám (2xy krát e umocněné na (x krát
y na druhou) plus 1) krát y s čárkou.
Měli jsme tolik y s čárkou
a přidali jsme další y s čárkou,
takže teď máme tolik
a ještě plus 1 y s čárkou.
Tohle se rovná...

Thai: 
มันจะเท่ากับ -- ผมตั้งใจ
บวก y ไพรม์ทั้งสองข้าง แล้วเราเหลือ 1
ลบ -- นี่คือพจน์เพี้ยนๆ -- y กำลังสองคูณ e
กำลัง xy กำลังสอง
และตอนนี้เราต้องหารทั้งสองข้างด้วยค่านี้
แล้วเราเหลือแค่อนุพันธ์ของ y เทียบ
กับ x เท่ากับค่านี้ ซึ่งผมจะลอกและวางลงไป
ที่จริง ขอผมเขียนมันใหม่นะ
 
เลื่อนลงมาหน่อย
มันเท่ากับ 1 ลบ y กำลังสอง
e กำลัง xy กำลังสองส่วนทั้งหมดนี้
ขอผมหาที่เพิ่มหน่อย
2xy e กำลัง xy กำลังสองบวก 1
เราก็เสร็จแล้ว
มันดูเพี้ยนๆ แต่มันไม่ได้ต่างไป
จากสิ่งที่เราทำมาในตัวอย่างก่อนๆ เลย

English: 
That's going to be equal
to-- well, I purposely
added y prime to both sides
and so we are left with 1
minus-- this is kind of a crazy
expression-- y squared times e
to the xy squared.
And now we just have to
divide both sides by this.
And we're left with the
derivative of y with respect
to x is equal to this, which
I will just copy and paste.
Actually, let me
just rewrite it.
Scroll down a little bit.
It's equal to 1
minus y squared e
to the xy squared
over this business.
Let me get some more space.
2xye to the xy squared plus 1.
And we're done.
It was kind of crazy, but
fundamentally no different
than what we've been doing
in the last few examples.

Czech: 
Schválně jsem k oběma
stranám přičetl y s čárkou,
takže zbyde 1 minus
tento šílený výraz,
a to (y na druhou) krát
e umocněné na (x krát y na druhou).
Nyní už jen musíme
obě strany vydělit tímhle.
Vyjde nám, že derivace y
podle x se rovná tomuhle,
což jen zkopíruji
a vložím sem...
Raději to přepíšu.
Rovná se to
1 lomeno...
Sjedu trochu níž.
Rovná se to 1 minus (y na druhou krát
e umocněné na (x krát y na druhou)),
to celé lomeno
tímhle výrazem,
tedy 2...
Budu potřebovat
víc místa.
...2 krát x krát y krát e umocněné na
(x krát y na druhou), tohle celé plus 1.
A máme hotovo.
Bylo to trochu šílené,
ale v zásadě nijak různé od toho, co jsme
dělali v několika předchozích příkladech.

Portuguese: 
Isso será igual - bem
eu adicionei de propósito
os y linha nos dois lados
e nos restou 1 menos-
essa é uma expressão meio
doida - y ao quadrado vezes
e elevado a x y ao quadrado.
Agora temos siplesmente que 
dividir ambos os lados por isso.
Ficamos com a derivada
de y com relação a x
que é igual a isso, eu vou 
copiar e colar.
Vou escrever
de novo.
Descendo mais a tela.
É igual a 1 menos y ao quadrado
vezes e elevado
a x y ao quadrado sobre isso aqui.
Vou arrumar mais espaço.
2xye elevado a x y ao quadrado mais 1.
Pronto, está feito!
Foi um pouco maluco mas
fundamentalmente nada diferente
do que vínhamos fazendo nos
outros exemplos.

Bulgarian: 
Това ще бъде равно на...Е, целенасочено
добавих y' към двете страни 
на уравнението, за да остане 1
минус... ето този вид щур израз
 y на квадрат по e
на степен x по y на квадрат.
Сега просто следва да разделим 
двете страни на този израз.
И остава производната на y спрямо x,
която е равна на този израз, който
 току-що копирах и поставих.
Всъщност нека да го запиша отново.
Равно е на 1 върху...
Слизам малко надолу.
y' е равно на 1 минус y^2 по e
на степен xy^2, върху ето този израз.
Нека да направя още малко място.
2xy по е на степен x по y^2 плюс 1.
И ето, че сме готови.
Беше много щуро, но принципно
не е по-различно
от това, което сме правили
 в предните няколко примера.

Korean: 
이것은
우변에 있는
1－y²e^(xy²) 와 같습니다
1－y²e^(xy²) 와 같습니다
 
이제 y' 을 구해야 합니다
복사해서 붙여넣겠습니다
그냥 다시 적겠습니다
 
화면을 조금만 내리겠습니다
1－y²e^(xy²)
1－y²e^(xy²) 를 좌변의
주황색 식으로 나눠야 합니다
공간이 조금 더 필요할 것 같습니다
2xye^(xy²)＋1 로 나눠주면 됩니다
이제 끝났습니다
매우 힘들었지만 근본적으로는
앞서 했던 것들과 
다른 것은 없었습니다

Portuguese: 
Legendado por Maria Oberlander
Revisado por Sérgio Fleury
