
Czech: 
V tomto videu zkusíme
zderivovat sinus z ln(x na druhou).
Máme tedy složenou funkci,
která je složená z další složené funkce.
Můžeme se na to dívat tak,
že když se f(x) rovná sin(x),
g(x) je přirozený logaritmus z x
a h(x) se rovná x na druhou,
tak se tohle rovná derivaci podle x z
funkce f v bodě (g v bodě h(x)).
Chtěl bych teď ukázat,
jak bych to dělal ve své hlavě,
aniž bych vypisoval celý
vzorec pro derivaci složené funkce.

English: 
So now we're going
to attempt to take
the derivative of the sine of
the natural log of x squared.
So now we have a function that's
the composite of a function,
that's a composite
of another function.
So one way you
could think of it,
if you set f of x as being equal
to sine of x, and g of x being
the natural log of x, and
h of x equaling x squared.
Then this thing right over
here is the exact same thing
as trying to take the
derivative with respect
to x of f of g of h of x.
And what I want to
do is kind of think
about how I would
do it in my head,
without having to write all
the chain rule notation.
So the way I would
think about this,

Thai: 
 
ตอนนี้เรากำลังพยายามหา
อนุพันธ์ของไซน์ของล็อกธรรมชาติของ x กำลังสอง
ตอนนี้เรามีฟังก์ชันที่เป็นการประกอบของฟังก์ชัน
ที่เป็นการประกอบของฟังก์ชันอีกตัว
วิธีคิดได้อย่างหนึ่ง
ถ้าคุณให้ f ของ x เท่ากับไซน์ของ x และ
g ของ x เท่ากับ
ล็อกธรรมชาติของ x และ
h ของ x เท่ากับ x กำลังสอง
แล้วอันนี้ตรงนี้จะเท่ากับ
การหาอนุพันธ์เทียบกับ
x ของ f ของ g ของ h ของ x
 
และสิ่งที่ผมอยากทำคือคิด
ว่าผมจะคิดในใจ
โดยไม่ต้องเขียนกฎลูกโซ่ออกมาได้อย่างไร
วิธีที่ผมจะคิด

Korean: 
함수 sin(ln(x²))의 도함수를 구해봅시다
함수 sin(ln(x²))의 도함수를 구해봅시다
이 함수는 합성 함수 내부에
합성 함수가 있는 함수입니다
그러므로 우리는 이것을 풀기 위하여
sin(x)를 f(x)로 두고  ln(x)를 g(x)
x²을 h(x) 로 둘 것입니다
그러면 구하려는 도함수는
f(g(h(x)))의 도함수와 같습니다
f(g(h(x)))의 도함수와 같습니다
 
우선 연쇄 법칙을 모른다고 하고
우선 연쇄 법칙을 모른다고 하고
이것을 어떻게 풀지 생각해봅시다
먼저

Bulgarian: 
Сега ще се опитаме да намерим
производната на синус от натурален
логаритъм от х^2.
Дадена е функция, която е съставна.
Тоест съставна е на друга функция.
Един начин, по който да мислиш за това,
е, ако избереш f(x) да бъде равно 
на синус от х,
g(x) да е натурален логаритъм от х, 
а h(x) да бъде равно на х^2.
Тогава това нещо ето тук 
е точно същото нещо,
като да се опитваш да намериш 
производната спрямо х
на f(g(h(x))) (f от g, от h от х).
Сега искам да помисля 
как бих решил задачата в главата си,
без да се налага да записвам пълното
означение за верижното правило.
Начинът, по който бих разсъждавал
по задачата,

Portuguese: 
Agora vamos tentar tomar a derivada
de seno do logaritmo
natural de x ao quadrado.
Temos uma função que é a composta
de uma função, que é a
composta de outra função.
Uma forma que podemos pensar, se
determinarmos f de x como seno de x.
E g de x como logaritmo natural de x.
E, vamos ver, f, g, vamos chamar h de x
de x ao quadrado, e então
isso aqui é a mesma coisa
que tomar a derivada
com relação a x, de f de g,
de g de h de x.
E o que quero fazer é, pense
sobre como eu iria fazer
na minha cabeça, sem ter que escrever
toda a notação da regra da cadeia.

Czech: 
Kdybych to dělal v hlavě,
tak bych se na to díval tak,
že jde o derivaci této vnější funkce f
podle celé složené funkce uvnitř.
Derivace sin(x) je cos(x),
ale místo kosinu v bodě x to bude
kosinus toho, co bylo tady uvnitř,
tedy kosinus z
přirozeného logaritmu...
Napíšu to tou
stejnou barvou.
...kosinus z ln(x na druhou).
x udělám tou
samou žlutou barvou.
Kosinus a tady bude
x na druhou.
Na tuhle část, kterou
jsem právě napsal,
se můžeme dívat jako na f s čárkou
v bodě (g v bodě h(x)).
Aby v tom bylo jasno, tak jsem zderivoval
vnější funkci podle toho, co bylo uvnitř.

Portuguese: 
O modo que iria pensar sobre isso, se
pensasse na minha cabeça,
seria a derivada da função
exterior de f com relação
ao nível de composição
diretamente abaixo dele.
Então a derivada de seno de x é cosseno
de x, mas ao invés de
ser cosseno de x será cosseno 
do que estava dentro dele.
Então será, cosseno do logaritmo natural.
Cosseno do logaritmo natural de x ao
quadrado, vou fazer o x em amarelo,
cosseno de x ao quadrado,
e você pode
realmente ver essa parte, o
que eu escrevi aqui, como
isso é f linha de g de h de x.
g de h de x, se quiser acompanhar
o que está acontecendo,
eu acabei de tomar a derivada
da função exterior com relação ao 
que estava dentro dela,

English: 
if I were doing this in
my head, is the derivative
of this outer function
of f, with respect
to the level of composition
directly below it.
So the derivative of
sine of x is cosine of x.
But instead of it
being a cosine of x,
it's going to be cosine of
whatever was inside of here.
So it's going to be
cosine of natural log--
let me write that
in that same color--
cosine of natural
log of x squared.
I'm going to do x that
same yellow color.
And so you could really view
this, this part what I just
read over here, as f prime,
this is f prime of g of h of x.
This is f prime of g of h of x.
If you want to keep
track of things.
So I just took the derivative
of the outer with respect
to whatever was inside of it.

Korean: 
가장 바깥쪽 함수인 f를
g(h(x))에 대하여
미분해 봅시다
sin(x)를 미분하면 cos(x)가 되죠?
그러므로 f(g(h(x)))를 미분하면
cos(g(h(x)))가 될 겁니다
즉 cos(ln(x²))이 되겠죠
즉 cos(ln(x²))이 되겠죠
즉 cos(ln(x²))이 되겠죠
 
이제 h(x)에 대해 똑같이 해봅시다
 
방금 적은 함수는
f`(g(h(x)) 로 표현할 수 있습니다
f`(g(h(x)) 로 표현할 수 있습니다
 
다시 한 번 말씀드리자면
저는 가장 바깥쪽 함수인 f(x)를
그 안의 요소에 대해서 미분한 겁니다

Bulgarian: 
ако правех това наум, е да търся производната
на тази външна функция от f спрямо
на съставните функции директно под нея.
Тоест производната на синус от х 
е косинус от х.
Но вместо да бъде равна на 
косинус от х,
ще бъде равна на косинус от това,
което се намира тук вътре.
Следователно ще бъде косинус 
от натурален логаритъм –
нека да го запиша в същия цвят –
косинус от натурален логаритъм от х^2.
Ще направя х в същия жълт цвят.
от x^2.
Наистина може да разглеждаш това, 
т.е. тази част, която току-що
прочетох ето тук, като f'. 
Това е f' от g, от h от х.
За да може да следиш решението.
Току-що намерих производната 
на външната функция спрямо
това, което се съдържа в нея.

Thai: 
ถ้าผมคิดในใจ คืออนุพันธ์
ของฟังก์ชันนอกคือ f เทียบกับ
ระดับการประกอบที่อยู่ข้างล่างมันพอดี
อนุพันธ์ของไซน์ของ x คือโคไซน์ของ x
แต่แทนที่จะเป็นโคไซน์ของ x
มันจะเป็นโคไซน์ของอะไรก็ตามที่อยู่ข้างในนี้
มันจึงเท่ากับโคไซน์ของล็อกธรรมชาติ --
ขอผมเขียนด้วยสีเดิมนะ --
โคไซน์ของล็อกธรรมชาติของ x กำลังสอง
 
ผมจะเขียน x ด้วยสีเหลืองสีเดิมนั้น
 
แล้วคุณก็มองอันนี้ ส่วนนี้ที่ผม
เพิ่งอ่านไปตรงนี้ ว่า f ไพรม์ นี่คือ
f ไพรม์ของ g ของ h ของ x
นี่คือ f ไพรม์ของ g ของ h ของ x
 
ถ้าคุณอยากตามสิ่งต่างๆ
ผมก็แค่หาอนุพันธ์ของตัวนอกเทียบกับ
สิ่งที่อยู่ข้างในมัน

Thai: 
และตอนนี้ผมต้องหาอนุพันธ์
ของตัวในเทียบกับ x
แต่ตอนนี้ผมมีฟังก์ชันประกอบอีกตัว
เราจะคูณอันนี้ด้วย เรา
จะใช้กฎลูกโซ่อีกครั้ง
เราจะหาอนุพันธ์ของล็อกเทียบกับ
x กำลังสอง
อนุพันธ์ของล็อกของ x เท่ากับ 1/x
แต่ตอนนี้เราจะได้ 1 ส่วน ไม่ใช่ x
แต่เป็น 1 ส่วน x กำลังสอง
 
ขอบอกให้ชัด ส่วนนี่ตรงนี้คือ g ไพรม์ ไม่ใช่ x
ถ้าเป็น g ไพรม์ของ x อันนี้จะเป็น 1 ส่วน x
แต่แทนที่จะเป็น x เรามี h ของ x ตรงนี้
เรามี x กำลังสอง
มันจะเท่ากับ g ไพรม์ของ x กำลังสอง
แล้วสุดท้าย เราหาอนุพันธ์
ของฟังก์ชันตัวในได้
ขอผมเขียนนะ
เราเขียนอันนี้ได้เป็น g ไพรม์ของ h ของ x
 
แล้วสุดท้าย เราแค่ต้องหา
อนุัพนธ์ของฟังก์ชันตัวในสุดเทียบกับ x

Korean: 
이제부터는 g(x)를 그 안의 요소에 대해 미분할 겁니다
이제부터는 g(x)를 그 안의 요소에 대해 미분할 겁니다
하지만 g(h(x))도 합성합수입니다
그래서 우리는 연쇄 법칙을 이용하여
이것들을 미분 할 겁니다
ln을 x²에 대해서 미분해봅시다
ln을 x²에 대해서 미분해봅시다
ln(x) 를 미분하면 1/x 가 됩니다
그러므로 ln(x²) 를 미분하면
1/x²이 됩니다
 
다시 명확하게 하자면 이 부분은
g`(x)가 아닙니다
x 대신에 h(x) 즉 x²을 넣어야 합니다
x 대신에 h(x) 즉 x²을 넣어야 합니다
그러므로 g`(x²)가 되죠
마지막으로
가장 안쪽 함수를 미분해 봅시다
다시 써보면
이 부분은 g`(h(x))가 됩니다
 
이제 마지막으로
가장 안쪽 함수인 h(x)를 
x에 대해 미분해봅시다

Czech: 
Nyní musím zderivovat
vnitřek podle x,
ale to je další
složená funkce,
takže tohle vynásobíme, opět podle
vzorce pro derivaci složené funkce,
derivací tohoto přirozeného
logaritmu podle (x na druhou).
Derivace přirozeného
logaritmu z x je 1 lomeno x,
ale my zde budeme mít nikoliv
1 lomeno x, ale 1 lomeno (x na druhou).
Aby to opět bylo jasné,
tak tato část je g s čárkou v...
Ne v bodě x, protože kdyby šlo o g(x)
s čárkou, tak by tu bylo 1 lomeno x,
ale místo x tu máme h(x)
neboli x na druhou,
takže je to g s čárkou
v bodě (x na druhou).
Nakonec zderivujeme
naši vnitřní funkci...
Tohle napíšu jako
g s čárkou v bodě h(x).
Nakonec zderivujeme funkci
nejvíc vevnitř podle x,

Bulgarian: 
Сега следва да намеря производната
на вътрешната функция спрямо х.
Сега обаче разполагаме с друга
 съставна функция.
Тоест ще умножим това или ще
приложим отново верижното правило.
Ще намерим производната на 
натурален логаритъм спрямо х^2.
Производната на натурален 
логаритъм от х е равна на 1/х.
Сега обаче няма да е равна 
на 1/х, а на 1/х^2.
1/х^2.
И така, за да изясним, тази част 
ето тук е g', но не от х.
Ако беше g'(х), това щеше да бъде равно на 1/х.
Но вместо х, имаме функцията h(х).
Дадено е х^2.
Тоест това е функцията g' от х^2.
И накрая, може да намерим 
производната на
вътрешната функция.
Нека да го запиша.
Може да запишем това 
като g' от h от х.
И накрая, просто следва да намерим
производната на най-вътрешната
 функция спрямо х.

English: 
And now I have to
take the derivative
of the inside with respect to x.
But now we have another
composite function.
So we're going to
multiply this times, we're
going to do the
chain rule again.
We're going to take the
derivative of ln with respect
to x squared.
So the derivative
of ln of x is 1/x.
But now we're going
to have 1 over not x,
but 1 over x squared.
So to be clear, this part right
over here is g prime of not x.
If it was g prime of x,
this would be 1 over x.
But instead of an x, we
have our h of x there.
We have our x squared.
So it's g prime of x squared.
And then finally, we
can take the derivative
of our inner function.
Let me write it.
So we could write this
is g prime of h of x.
And finally, we
just have to take
the derivative of our innermost
function with respect to x.

Portuguese: 
e agora eu tenho que tomar a derivada do
interior em relação a x.
Mas agora temos outra função
composta, então temos
que multiplicar isso por, vamos usar
a regra da cadeia de novo.
A derivada de, nós vamos tomar a
derivada do logaritmo com
relação a x ao quadrado.
Então, a derivada de ln de x
é um sobre x, mas agora
vamos ter um sobre, não x, mas
um sobre x ao quadrado.
Um sobre x ao quadrado.
Para ficar claro, essa parte aqui, essa
parte aqui é g linha, não de x.
Se fosse g linha de x, seria um sobre x.
Mas ao invés de x, nós
temos nosso h de x aqui.
Temos o x ao quadrado.
Então é g linha de x ao quadrado.
E finalmente, podemos tomar a derivada
de nossa função interior.
Deixe-me escrever.
Podemos escrever esse g linha de h de x.
E finalmente, só temos que tomar a
derivada de nossa função
interior com relação a x.

Thai: 
อนุพันธ์ของ 2x เทียบกับ --
อนุพันธ์ของ x กำลังสองเทียบกับ x เท่ากับ 2x
คูณ h ไพรม์ของ x
ขอผมทำให้ทุกอย่างชัดเจนนะ
สิ่งที่เรามีตรงนี้สีม่วง
อันนี้ อันนี้ และอันนี้คือสิ่งเดียวกัน
พจน์หนึ่งเขียนออกมาชัดเจน 
พจน์หนึ่งเขียนแบบนามธรรม
อันนี้ อันนี้ และอันนี้คือสิ่งเดียวกัน
เขียนแบบชัดเจน กับเขียนแบบนามธรรม
แล้วสุดท้าย อันนี้ อันนี้ และอันนี้คือสิ่งเดียวกัน
เขียนแบบชัดเจน กับเขียนแบบนามธรรม
แต่เราเสร็จแล้ว
ที่เราต้องทำที่เหลือก็แค่จัดรูปให้ง่ายลง
ถ้าเราเปลี่ยนลำดับการคูณ
เราได้ 2x ส่วน x กำลังสอง
ผมก็ตัดได้
x ส่วน x นี่, 2x ส่วน x กำลังสอง
เท่ากับ 2 ส่วน x
และเรากำลังคูณทั้งหมดด้วยตัวนี้

Korean: 
x²을 x에 대해 미분하면
2x가 됩니다
그러므로 h`(x)를 곱합니다
처음부터 끝까지 정리해보겠습니다
보라색으로 표시된 부분들은
똑같은 함수들입니다
위의 식은 명확하게 
밑의 식은 치환하여 표현했습니다
이 부분들도 하나는 명확하게
하나는 치환하여 표현된 같은 함수입니다
마찬가지로 이 부분들도
같은 함수입니다
이제 거의 끝났습니다
이제 우리가 해야할 것은
이것들을 단순히 곱하는 것입니다
2x와 1/x²를 곱해봅시다
약분을 해 보면
2x × 1/x²은
2/x와 같습니다
이런 식으로 모두 곱합니다

Portuguese: 
A derivada de duas vezes x em relação a,
a derivada de x ao quadrado com
relação a x, é duas vezes x.
Então vezes h linha de x.
Deixe-me deixar claro.
Isso, deixe-me ver.
O que temos aqui em roxo,
isso, isso e isso são
as mesmas coisas, uma expressa de forma
concreta, uma expressa de forma abstrata.
Isso, isso e isso, são a mesma coisa,
expressas de forma concreta e abstrata.
E finalmente, isso e isso são
a mesma coisa, expressas
de forma concreta e abstrata.
Então, acabamos.
Tudo o que temos que fazer para
terminarmos é simplificar isso, então se
mudarmos a ordem da multiplicação, temos
duas vezes x sobre x ao quadrado,
então podemos cancelar.
Esse x sobre x, dois x sobre x ao
quadrado é a mesma coisa
que dois sobre x, e estamos
multiplicando por todo o resto.

Bulgarian: 
Тоест производната на 2х спрямо...
или производната на х^2 спрямо х,
 е равна на 2х.
Така че умножаваме по h' от х.
Нека да изясня всичко.
Какво имаме ето тук, което 
е записано в лилаво?
Това, това и това са едно и също нещо.
Едното е представено конкретно, 
а другото абстрактно.
Това, това и това са едно и също нещо,
представени конкретно и абстрактно.
И накрая, това, това и това 
са едно и също нещо.
представено конкретно и абстрактно.
Тогава сме готови.
Всичко, което следва да направим, е 
да опростим получения израз.
Може просто променим реда, 
по който умножаваме.
Имаме 2х върху х^2,
така че мога да извърша съкращение.
Това х върху х...
 т.е. 2х върху х на квадрат,
е същото нещо като 2 върху х.
И го умножаваме по целия този израз.

Czech: 
což je derivace z
(2 krát x) podle x...
Bude to derivace z (x na druhou)
podle x, což je 2 krát x,
tedy krát h s čárkou
v bodě x.
Ještě to ujasním.
Tohle...
To, co teď zvýrazňuji fialově,
se sobě rovná.
Jednou je to napsané konkrétně,
podruhé obecně.
Tyhle dva výrazy se
sobě také rovnají,
jednou je to napsané konkrétně,
jednou obecně.
A nakonec i tyto dva výrazy
se sobě rovnají,
jen je to jednou napsáno
konkrétně a jednou obecně.
Tím máme hotovo,
respektive abychom měli hotovo,
musíme tohle už jen zjednodušit.
Když změníme pořadí,
ve kterém násobíme,
tak dostaneme (2 krát x)
lomeno (x na druhou),
takže můžeme krátit.
x lomeno x...
(2 krát x) lomeno (x na druhou)
je to samé co 2 lomeno x,
a to ještě násobíme
celým tímhle výrazem.

English: 
So the derivative of
2x with respect to--
or the derivative of x squared
with respect to x is 2x.
So times h prime of x.
Let me make everything clear.
So what we have right
over here in purple,
this, this, and this
are the same things.
One expressed concretely,
one expressed abstractly.
This, this, and this
are the same thing,
expressed concretely
and abstractly.
And then finally, this and
this are the same thing,
expressed concretely
and abstractly.
But then we're done.
All we have to do to be done
is to just simplify this.
So if we just change the order
in which we're multiplying,
we have 2x over x squared.
So I can cancel some out.
So this x over x,
2x over x squared
is the same thing as 2 over x.
And we're multiplying it
times all of this business.

Thai: 
เราจึงเหลือ 2 ส่วน x
อันนี้หายไป
2 ส่วน x คูณโคไซน์ของ
ล็อกธรรมชาติของ x กำลังสอง
มันดูเหมือนอนุพันธ์ที่น่าเหนื่อยใจ
แต่เราแค่บอกว่า โอเค อนุพันธ์ของไซน์
ของอะไรสักอย่าง
เทียบกับอะไรสักอย่างนั้นคืออะไร?
มันก็คือโคไซน์ของอะไรสักอย่างนั้น
แล้วเราเข้าไปชั้นหนึ่ง อนุพันธ์
ของอะไรสักอย่างนั้นคืออะไร?
ในอะไรสักอย่างนั่น เรามีการประกอบอีกตัว
อนุพันธ์ของ ln ของ x หรือ ln ของอะไรสักอย่าง
เทียบกับอะไรสักอย่างอีกตัว มันคือ
1 ส่วนอะไรสักอย่างนั้น
แล้วเราได้ 1 ส่วน x กำลังสองตรงนี้
กำลังสองนั้นตัดไป
แล้วสุดท้าย อนุพันธ์ของฟังก์ชันในสุดนี้
มันเหมือนกับการปอกหัวหอม
อนุพันธ์ของฟังก์ชันใน
เทียบกับ x ซึ่งก็คือแค่ 2x
ซึ่งเราได้ตรงนี้
นี่คือ 1 ส่วน x กำลังสอง
นี่คือ 2x ก่อนที่เราจะตัดมันไป
หวังว่ามันคงช่วยให้กระจ่างขึ้นนะ

Portuguese: 
Restou dois sobre x, isso é cancelado.
Dois sobre x vezes o cosseno do logaritmo
natural de x ao quadrado.
Parecia uma derivada
assustadora, mas dissemos ok,
qual é a derivada de seno de
algo com relação a esse algo
Bom, é cosseno desse algo.
E então vamos para outra camada.
Qual a derivada desse algo?
Bom, nesse algo, temos
outra composição.
A derivada do logaritmo de x, ou logaritmo
de alguma coisa com relação a outra coisa.
Isso será um sobre a coisa,
nós obtivemos um sobre x
ao quadrado, o quadrado foi cancelado,
e então encontrar a derivada
dessa função interior.
É como descascar uma cebola.
A derivada da função
interior em relação a x,
que é duas vezes x, que obtivemos aqui.
Isso era um sobre x ao quadrado,
isso era dois x antes de cancelar.
Espero que tenha esclarecido um pouco.
[Traduzido por: Victória Celeri]

Czech: 
Zbylo nám tedy
2 lomeno x...
Tohle se
pokrátilo.
...(2 lomeno x) krát
kosinus z ln(x na druhou).
Vypadalo to sice jako
strašidelná derivace,
ale nejdřív nás jen zajímalo, jak vypadá
derivace sinu něčeho podle toho něčeho,
což je kosinus
toho něčeho.
Pak se posuneme o jednu vrstvu dovnitř
a ptáme se, co je derivace toho něčeho,
přičemž naše něco byla
opět složená funkce,
takže derivace z ln(x)...
Derivace přirozeného logaritmu něčeho
podle toho něčeho je 1 lomeno to něco,
takže nám tady vyšlo
1 lomeno (x na druhou),
což se nám
později pokrátilo.
Nakonec jsme museli spočítat
derivaci funkce nejvíce uvnitř.
Je to jak loupat cibuli.
Museli jsme zjistit derivaci této
vnitřní funkce podle x, což je 2 krát x.
Máme to tady.
Tady bylo 1 lomeno (x na druhou) a tady
2 krát x předtím, než jsme pokrátili.
Snad v tom teď
máte trochu víc jasno.

Bulgarian: 
И така, остана 2 върху х.
Ето това изчезва.
2/х по косинус от натурален 
логаритъм от х на квадрат.
Изглеждаше много обезсърчително 
да търсим производната.
Но просто си казахме : Да намерим 
производната на синус от нещо
спрямо това нещо.
Е, това е косинус от нещо.
След това навлизаме едно ниво
навътре. Тоест: на какво е равна
производната на ето това нещо?
Е, в това нещо имаме друга съставка.
Производната на натурален логаритъм от х...
 или натурален логаритъм от нещо спрямо
друго нещо...
ще бъде равна на 1 върху нещото.
Получихме 1 върху х на квадрат тук,
защото този квадрат се съкращава.
И накрая, производната на тази 
най-вътрешна функция.
Процесът прилича на обелване на лук.
Производната на тази вътрешна функция
спрямо х, което беше равно 
просто на 2х.
Което получихме ето тук.
Това беше 1/х^2.
Това беше 2х преди 
да извършим съкращението.
Надявам се, че това ще помогне, 
за да успееш да си изясниш нещата.

English: 
So we're left with 2 over x.
And this goes away.
2 over x times the cosine of
the natural log of x squared.
So it seemed like a very
daunting derivative.
But we just say, OK, what's the
derivative of sine of something
with respect to that something?
Well, that's cosine
of that something.
And then we go in
one layer, what's
the derivative of
that something?
Well, in that something we
have another composition.
So the derivative of ln of x,
or ln of something with respect
to another something,
well that's
going to be 1 over
the something.
So we had gotten a 1
over x squared here,
that squared got canceled out.
And then finally, the derivative
of this innermost function,
it's kind of like
peeling an onion.
The derivative of
this inner function
with respect to x,
which was just 2x.
Which we got right over here.
This was 1 over x squared.
This was 2x before we
did any canceling out.
So hopefully that helps
clear things up a little bit.

Korean: 
그 결과는
그 결과는
cos(ln(x²)) × 2/x 입니다
복잡한 미분처럼 보이지만
하나씩 다시 해 봅시다
sin 함수를 미분하면
cos 이 됩니다
그럼 이제 함수 내부의
다른 함수를 미분해봅시다
그 함수 안에는 다른 함수가 있습니다
즉 안쪽 함수의 자연로그 값을
안쪽 함수에 대해 미분하면
1 나누기 안쪽 함수가 될 겁니다
그러므로 1/x²을 우리가 적은 것이죠
약분했지만 원래는 제곱이었죠?
마지막으로 가장 안쪽 함수를 미분합니다
양파 껍질을 벗기는 것처럼요
이 안쪽 함수를 x에 대해 미분하면
2x 입니다
여기에 방금 적었듯이
이 부분은 1/x² 이었고
이 부분은 2x였죠
조금이나마 도움이 되었다면 좋겠습니다
