help-10

[normal spindle]

2√5/5?

wait actually

-6√5/25 - -4√/5/25 would evaluate to what

[autumn gale]

double negative makes a positive

so it would be -6√5 +4√5

[normal spindle]

so

i put -2√5/25

and its still wrong

now i am confused

[obtuse pebble]

@normal spindle Has your question been resolved?

[normal spindle]

<@&286206848099549185>

this qs

i got 19√3/26 and its wrong

did the whole formula n all

sin a is -12/13 cos a is -5/13

sin b is -1/2 and cos b is -√3

cos(a) cos(b) - sin(a) sin(b)

ping me back pls

[obtuse pebble]

@normal spindle Has your question been resolved?

[obtuse pebble]

@normal spindle Has your question been resolved?

[obtuse pebble]

Channel closed

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[lone bolt]

what do i do next?

[fierce lagoon]

Do you have to use the limit definition

[lone bolt]

for this yes

[high lily]

would you be able to simplify
$$\frac{\br{\frac ab}}{c}$$

[warm shale]

ℝamonov

[lone bolt]

a/c(b)

[high lily]

no

still no

[lone bolt]

how about now

[high lily]

still no

[lone bolt]

1a/bc

🏳️

[brisk matrix]

$\frac{(\frac{a}{b})}{c} = \frac{(\frac{a}{b})}{(\frac{c}{1})}$

[warm shale]

maximo

[brisk matrix]

if this is meant to be a/(bc) then yes

[lone bolt]

a/c(b) is the same thing

l0l?

[brisk matrix]

that looks like youre saying (a/c) * (b)

[lone bolt]

were nat calculators tho

not

u knew what i meant? xd

[brisk matrix]

well no

[high lily]

no, its 50:50

[lone bolt]

ok

a/(bc)

then

[high lily]

hence why proper use is important

apply that to what you have

also

[lone bolt]

or a/(c(b))

[high lily]

that radical line should clearly indicate that 7 is under it

a/(c(b))
yes but a bit excessive now

[lone bolt]

it is

mb

[high lily]

its supposed to

but your line falls short

[lone bolt]

how are my + signs

are they up to par

😄

[high lily]

passable

[lone bolt]

woo

[high lily]

understandable that they're + signs in context

first is a bit slanted,
second is missing the top
third is a bit faint

[lone bolt]

got it, will fix be more mindful from now on

but putting that to the side

whats next

[high lily]

,rotate

[lone bolt]

thats cool

😎

[high lily]

factor out 3 for convenience

then use conjugates to rationalise the numerator

(do not explicitly expand the denominator)

[lone bolt]

3[-3(.... etc)?

conjugates ?

[high lily]

write the full thing

yeh.

you should've been introduced to that if you're being asked to to this

[lone bolt]

took precalc years ago

i have to relearn some stuff

[high lily]

have you ever been asked to simplify stuff like
$$\frac{1}{2-\sqrt{3}}$$
and/or recall how to do it

[warm shale]

ℝamonov

[lone bolt]

o.o

i have no clue man? 1/2 * 3^-(1/2)?

[high lily]

no

$\frac{a}{b-c} \redneq \frac ab \cdot \frac ac$

[warm shale]

ℝamonov

[lone bolt]

ah

well

-c

-b i mean

f idk

[high lily]

still wrong on multiple levels

[lone bolt]

cant the 2 borrow the negative? 😄

[high lily]

what

this is just nonsense at this point

[lone bolt]

ok so what should i do

to know how to simplify this

cause im struggling

[high lily]

in short,
a + b and a - b
are conjugates of each other

[lone bolt]

yeah

opposites

[high lily]

do NOT use the word oppopsite here

opposite is so vague

[lone bolt]

ok

[high lily]

use the actual mathematically unambiguously defined term which is conjugates

to rationalise the denominator of $\frac{1}{2-\sqrt{3}}$ you can multiply the numerator and denominator by the conjugate of the denominator.
the conjugate of $2 - \sqrt{3}$ is $2 + \sqrt{3}$
$$\frac{1}{2-\sqrt{3}} = \frac{1}{2-\sqrt{3}} \cdot \frac{2+\sqrt{3}}{2+\sqrt{3}}$$

[warm shale]

ℝamonov

[high lily]

note that the conjugate is used due to the factorisation difference of two squares

(a-b)(a+b) = a^2 - b^2,
doing this will rid yourself of square roos in the denominator

[lone bolt]

ok so u multiply by conjugate to num and denom so it cancels out in the denom

ok

[high lily]

wdym by cancels in the denom

[lone bolt]

=1

lol

[high lily]

isn't cancelling

[lone bolt]

square root has + and -, u found conjugate which made it = to 1 in the denom

[high lily]

ugh

very poor wording

(2 - sqrt(3))(2+sqrt(3)) = 2^2 - 3 = 1

[lone bolt]

ok

[high lily]

no cancellation here

[lone bolt]

so whats that called then

in your words

[high lily]

applying factorisation identity for difference of two squares / simplification

[lone bolt]

gt it

got

so am i not factoring the -3

on the num

or simplifying denom first?

[high lily]

back to the question you have?

[lone bolt]

should i solve this first lol

[high lily]

yes,
so what's the end result after using conjugates and simlifying?

$$\frac{1}{2-\sqrt{3}} = \frac{1}{2-\sqrt{3}} \cdot \frac{2+\sqrt{3}}{2+\sqrt{3}} = , ?$$

[warm shale]

ℝamonov

[lone bolt]

2 + square root 3 😇

[high lily]

yeh

also just write sqrt(3)

[lone bolt]

squirt3

gotit

[high lily]

sqrt(3)

not squirt3

parentheses not optional either

[lone bolt]

💦 3

[high lily]

as they clearly indicate what's being square rooted

[lone bolt]

sqrt(3)

[high lily]

the end result is
2 + sqrt(3)
yes

[lone bolt]

am i certified by Ramonov to move on to the next step?

[high lily]

yes

[lone bolt]

excellent

what now

[lone bolt]

here

[high lily]

factor out 3 (from the numerator) for convenience

then apply the same idea to your question,
just instead of rationalising the denominator, you're rationalising the numerator
multiply the numerator and denominator of your fraction by the conjugate of the component you wish to rationalise

[lone bolt]

whats next math wizard

if i did it correctly lol

i have my eraser ready

[high lily]

you didn't actually use the conjugate

the conjugate of
a - b
is NOT
-a + b

[lone bolt]

thought there were two sqrts

but alright

so i erase the - in front then

[high lily]

a and b just represent the first and second terms, whatever they may be

so i erase the - in front then
yes

and you need to fix the simplification after that as well

[lone bolt]

yes

3 negatives

i erased

[high lily]

$a^2 - b^2$ isn't 1 for all $a,b$

[warm shale]

ℝamonov

[lone bolt]

a^(1/2) you mean?

[high lily]

i mean what i said

[lone bolt]

where is a 2nd power

[bleak rain]

hi guys
could you recommend me the best pages for mathematical olympics please

[lone bolt]

#help-20 @bleak rain

[high lily]

how are you getting just
-3 for
$$-3(\sqrt{x+7} - \sqrt{x+h+7})(\sqrt{x+7} + \sqrt{x+h+7})$$

[warm shale]

ℝamonov

[lone bolt]

because they = 1

and u told me

[high lily]

no

[lone bolt]

to simplify

find conjugate

[high lily]

multiplying conjugates together doesn't always give you 1

it was 1 in the previous example because

(2 - sqrt(3))(2+sqrt(3)) = 2^2 - 3 = 1

[lone bolt]

ohok

[high lily]

where is a 2nd power
the square / power of two comes from the factorisation identity for a difference of two squares

note that the conjugate is used due to the factorisation for the difference of two squares
(a-b)(a+b) = a^2 - b^2

[lone bolt]

so am i left with 3(x+7)^2/2 or 3(x+7) by itself?

[high lily]

neither

[warm shale]

ℝamonov

[lone bolt]

a = sqrt(x+7 ) , b=sqrt(x+h+7)

[high lily]

yes

and applying the identity
$$(a-b)(a+b) = a^2 - b^2$$
can you simplify the part in red

[warm shale]

ℝamonov

[lone bolt]

so i do this to the denom too

well

[high lily]

no

[lone bolt]

denom is different

[high lily]

focussing on the red

and as mentioned earlier, do NOT expand the denominator

can you simplify the part in red
do NOT worry about anything else

at least for now

[lone bolt]

3[(sqrt (x+7)^2) - (sqrt(x+h+7)^2)]

[high lily]

simplify that further

[lone bolt]

3 [(x+7)-(x+h+7)]

[high lily]

yeh, simplify further

[lone bolt]

3[h]

[high lily]

no

[lone bolt]

3-h

[high lily]

no

[lone bolt]

-3h

[high lily]

yes

[lone bolt]

time for the denom?

[high lily]

and as mentioned earlier, do NOT expand the denominator

[lone bolt]

trying to find h'(x) though

[high lily]

yeh and i stand by what i said

[lone bolt]

ok so let me write down

the final answer

minus the h-> 0

[high lily]

after simplifying the numerator, the difference quotient becomes
$$\frac{-3\red{h}}{\red{h}\sqrt{x+7}\sqrt{x+h+7}(\sqrt{x+7} + \sqrt{x+h+7})}$$

[warm shale]

ℝamonov

[lone bolt]

ok

[high lily]

the factors of h, marked in red cancel

and you can now take the limit as h→0 to get the derivative without getting an indeterminate form

[lone bolt]

thought they equal 1

[high lily]

wdym by they

[lone bolt]

h/h

=1

[high lily]

h/h = 1 yes

[lone bolt]

so they cancel

when u want them to

but not when i want them to

😄 seems fair

[high lily]

the factors of h, marked in red cancel

[lone bolt]

so the denom doesnt cancel? or =1, which one is it ramonov

[high lily]

wdym

the factors of h cancel

that's all that cancels

the same way 6/9 simplifies to 2/3 from cancelling a factor of 3 in the numerator and denomiantor

[lone bolt]

alright so youre left with -3/ all of that

[lone bolt]

here

now plug in 0 for h

for h'(x)?

i got -3/[sqrt(x+7)^2 ((sqrt(x+7) + sqrt(x+7))]

[lone bolt]

fair point

it simplifies out

[high lily]

um did you go backwards or something and/or did stuff yuo weren't supposed to

what happened to the - sign
and why do you stuff have h after plugging in 0 for h

[lone bolt]

ok check now

[high lily]

ok, simplify that

[lone bolt]

https://tenor.com/view/yes-drill-sergeant-yes-boss-yes-sir-all-right-gif-22599540

((x+7) (2sqrt(x+7))?

[high lily]

no

[lone bolt]

sqrt (x+h) + sqrt (x+h) doesnt = 2 sqrt (x+h)?

[high lily]

it does

[lone bolt]

so whats wrong

and sqrt (x+7) ^ 2 = x+7

[high lily]

you wiped the numerator from existence

[lone bolt]

well

of course

we were focued

focused on denom

[high lily]

how was i supposed to know what you were focussing on

[lone bolt]

-3/ ((x+7) (2sqrt(x+7))

[high lily]

i got -3/[sqrt(x+7)^2 ((sqrt(x+7) + sqrt(x+7))]

ok, simplify that
that refers to the whole thing

that (whole thing in case I'm not being clear enough) can be expressed more nicely

[lone bolt]

well i cant use conjugate

and (2 sqrt (x+7)) is trapped

in the parenthesis

so im screwed

doesnt -3/[sqrt(x+7)^2 ((sqrt(x+7) + sqrt(x+7))] = -3/ ((x+7) (2sqrt(x+7))

@high lily did you give up on your student

[high lily]

yeh

it does

i didn't say there was anything invalid going on there
$$-\frac{3}{(x+7)2\sqrt{x+7}}$$
\verb|can be expressed more nicely|

[warm shale]

ℝamonov

[lone bolt]

alright

any further steps?

[high lily]

express it more nicely

[lone bolt]

to find h'(x)?

[high lily]

you already have something that's equivalent to h'(x)
the rest is asthetics

would you agree that something like
$$\frac{3}{a5b}$$
looks ugly af?

[lone bolt]

-3/(2(x+7)(sqrt(x+7))

[warm shale]

ℝamonov

[high lily]

-3/(2(x+7)(sqrt(7))
missing an x

-3/(2(x+7)(sqrt(x+7))
yes, alright now

[lone bolt]

done?

[high lily]

yes

[lone bolt]

so i can pop open the champane?

and make it sqrt?

squirt

[high lily]

if you're of legal drinking age, sure

[lone bolt]

mathway says that the answer

is -3/ (2(x+7)^(3/2))

?

the derivative of 3/(sqrt(x+7))

[high lily]

please stop making typos

[lone bolt]

i cant speak calculator

[high lily]

the derivative of 3/(sqrt(x+h))

[lone bolt]

im native

[high lily]

you mean
3/(sqrt(x+7))

[lone bolt]

yes

my bad

[high lily]

there are multiple different ways to express something depending on personal preference

$k\sqrt{k} = k^\frac32$

[warm shale]

ℝamonov

[high lily]

either form would be acceptable

[lone bolt]

so you mean to tell me you can change this into -3/ (2(x+7)^(3/2))

[high lily]

yes

if you want

[lone bolt]

-3/(2(x+7)(sqrt(x+7)) = -3/(2(x+7)((x+7)^(1/2))?

[high lily]

missing (

yes

think you have an extra )

[lone bolt]

-3/(2(x+7)((x+7)^(1/2)) = -3/ (2(x+7)^(3/2)) how exactly?

[high lily]

exponent laws

if you insist,
(x+7) = (x+7)^1

you don't have any issue with stuff like
3 * 3 = 3^2 do you?

[lone bolt]

no

i have no beef

with exponents

we're on good terms

for now

[high lily]

so what is

how exactly?
about

[lone bolt]

continue 🙂

huh

[high lily]

that implied you had an issue with

-3/(2(x+7)((x+7)^(1/2)) = -3/ (2(x+7)^(3/2))
what's your issue with that

[lone bolt]

lol

thats so nasty

what about the 2 in front

it doesnt get any love?

doesnt the parenthesis mean anything

l0l

so much strange manipulation

if only math could talk

nv

nvm

the 2 is outside

lmao

ok

i understand

but let me ask you

why did u take out the 2

in the first place

one step before

[lone bolt]

here

like what rule is that

thats so dirty

very inappropriate

you took a number that was in being multiplied by sqrt and just took it out back

[high lily]

commutative property of multiplication

a * b * c = c * a * b

(x+7) and sqrt(x+7) are just expressions

what's being done/applied there is no different from expressing
$$\frac{3}{a5b}$$
as
$$\frac{3}{5ab}$$

[warm shale]

ℝamonov

[lone bolt]

the first one?

[high lily]

second one

[lone bolt]

there arent parentheses though

[high lily]

doesn't matter

a,b just represent expressions that are being multiplied together whatever they may be

[lone bolt]

sure it isnt the first lol

2 is multipled by the sqrt (x+7) tho

nothing else?

😄

[high lily]

all three terms are being multiplied together

explicitly indicating the multiplication
$$\underbrace{(x+7)}{a} \cdot \underbrace{2}{b} \cdot \underbrace{\sqrt{x+7}}_{c}$$

[warm shale]

ℝamonov

[high lily]

$=\underbrace{2}{b} \cdot \underbrace{(x+7)}{a} \cdot \underbrace{\sqrt{x+7}}_{c}$

[warm shale]

ℝamonov

[lone bolt]

you cant combine a and c tho

c is sqrt

and a isnt

l0l

[high lily]

wdym

[lone bolt]

oh

nvm

u can

l0l

[high lily]

noones combining anything at this stage

[lone bolt]

.....

hahahaha

well i am

x+7 and sqrt x +7

[high lily]

rearranging the 2 is just the application of the commutative property

nothing else

[lone bolt]

= -3/(2(x+7)^(3/2))

[high lily]

if you wish to combine the (x+7) * sqrt(x+7),
that's an application of exponent/radical laws

[lone bolt]

if u type that into mathway

original function

and ask for derivative

it will be in that form

[high lily]

so what?

[lone bolt]

here

so thats the way

i want it

ramonov, its been a pleasure

ill make sure to improve my hand writing and math skills

youve been so helpful

https://tenor.com/view/kitten-love-pet-cat-cute-gif-13582175

ty

.close

[obtuse pebble]

Channel closed

Closed by @lone bolt

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cloud saffron]

how do i apply the chain rule here? the formula in my textbook doesn't seem to match up with how i am trying to apply it

ddx[f(g(x))] is f'(g(x))g'(x)

i don't know why i can't take the derivative of x^2 first then multiply by 2u... as the derivative of g(x)

is this incorrect?

but then i would end up with x^2 * y once i sub back in the value of g(x) which doesn't clear anything up. i still have a composite function there.

[trim locust]

You’re fine until you use a u sub, implicit differentiation says that $\frac{d}{dx} ( g(y) ) = g’(y) \frac{dy}{dx}$

[warm shale]

lewis_f04

[cloud saffron]

i did something wrong with implicit differentiation

i went about applying the chain rule then product rule

i lost my dy/dx

implicit differentiation is still not clear to me

[brisk matrix]

y is a function of x

so when you do $\dv{x}y^3$, you need to do the chain rule

[warm shale]

maximo

[brisk matrix]

this is in general to any function of y

[cloud saffron]

where is y^3 here? I see y^2

[brisk matrix]

third line

but again, even if it’s d/dx of y^2

you still need to apply the chain rule

y is a function of x

[cloud saffron]

ok

so chain rule takes precedence over the product rule?

[brisk matrix]

no

[cloud saffron]

i'm confused because i don't see y^3 as being a composite function

[brisk matrix]

each has their own time to be applied

f(y(x))

where f(x) = x^3

and y(x) is some function of y

so d/dx of f(y) = 3y^2 * y’

[cloud saffron]

sorry i'm lost

on the second line as i see y^3 alone i wouldn't think to apply the chain rule

with x^2y^2 yes

[brisk matrix]

i’m telling you you need to do chain rule because you’re deriving with respect to x

and y is a function of x

d/dy of y^3 = 3y^2

d/dx of y^3 = 3y^2 * y’

[cloud saffron]

oh i see

i'm placing y prime on the end of my derivative?

ddx[f(g(x))] is f'(g(x))g'(x)

is chain rule

[brisk matrix]

yes

f is x^3, g is y

[cloud saffron]

i thought i was deriving g(x) right there...

but i just place y prime on the end and continue?

[brisk matrix]

where

[cloud saffron]

d/dy of y^3 = 3y^2
d/dx of y^3 = 3y^2 * y’

[brisk matrix]

what about this

[cloud saffron]

i don't understand how y prime was added to the end here

[brisk matrix]

y^3 = f(g(x))

where f(x) = x^3

g(x) = y

[cloud saffron]

the notation of the chain rule

[brisk matrix]

before we get to the chain rule

[cloud saffron]

says i multiply by g prime of x

[brisk matrix]

do you understand what i’ve sent?

[cloud saffron]

oh ok

yes

[brisk matrix]

y^3 is a composition of functions

[cloud saffron]

yes

[brisk matrix]

so to take its derivative

we need the chain rule

[cloud saffron]

yes

i understand that

[brisk matrix]

what is g’(x) if g(x) = y

[cloud saffron]

1

[brisk matrix]

no

[cloud saffron]

derivative of a variable is 1

[brisk matrix]

no

derivative of x with respect to x is 1

the derivative of y(x) with respect to x is y’(x)

[cloud saffron]

that's where i'm lost

[brisk matrix]

what’s the derivative of f(x)?

[cloud saffron]

2x

[brisk matrix]

f’(x)

don’t worry about the x^2

just for an arbitrary function

the derivative of f(x) is f’(x)

[cloud saffron]

yes

[brisk matrix]

or df/dx

so the derivative of y(x)

is y’(x)

or dy/dx

it’s no different

[cloud saffron]

ok

can you write out the steps we just spoke about

?/

[brisk matrix]

write what

it’s just the chain rule

try it yourself

[cloud saffron]

one second. i'll edit what i've done

still unclear as when i'm differentiating with respect to x or y

when to use d/dx and when to use dy/dx

[brisk matrix]

you’re always differentiating with respect to x

remember the bottom is the differentiation variable

dy/dx means the derivative of y with respect to x

[cloud saffron]

yes

[brisk matrix]

it doesn’t mean we’re deriving with respect to y

[cloud saffron]

oh ok

i missed that

have i corrected my mistakes? it still seems unclear. i try applying the chain rule using the syntax given by the formula but the logic doesn't make sense to me yet

[brisk matrix]

there are still mistakes

let's work through a quick exaple

[cloud saffron]

ok

[brisk matrix]

$\dv{x}y = \dv{y}{x}$

[warm shale]

maximo

[brisk matrix]

we're assuming y is a function of x

so if we take the derivative of y with respect to x, we can just call it dy/dx

[cloud saffron]

yes

[brisk matrix]

now let's consider y^2

$\dv{x}(y^2) = \dv{x}f(g(x)) = f'(g(x))\cdot g'(x)$

[warm shale]

maximo

[brisk matrix]

where f(x) =x^2, g(x) = y(x)

that is

$\dv{x} (y^2) = 2(y(x))\cdot y'(x)$

[warm shale]

maximo

[brisk matrix]

do you see how i got to this

[cloud saffron]

how is y^2 a composition of functions?

[brisk matrix]

y^2 = (y)^2

so we're doing f(y)

and y is a function of x

[cloud saffron]

we have a variable raised to the second power

[brisk matrix]

no

again

y is a function

not just a variable

[cloud saffron]

i'm still unclear

[brisk matrix]

$\dv{x}(x)$ what is this?

[warm shale]

maximo

[cloud saffron]

the derivative with respect to x

[brisk matrix]

what does it compute to though

[cloud saffron]

i still want to say 1

[brisk matrix]

yes

how about $\dv{w}(x)$?

[warm shale]

maximo

[cloud saffron]

i don't know. that doesn't make sense to me

the derivative of x with respect to w

[brisk matrix]

yes

now if x is a function of w

then it's just that

dx/dw

but if x is constant with respect to w

then the derivative just becomes 0

kind of like how $\dv{x}(2) = 0$

[warm shale]

maximo

[brisk matrix]

or in general

$\dv{x}(c) = 0$

[warm shale]

maximo

[brisk matrix]

does that make sense?

[cloud saffron]

yes. from previous lessons

[brisk matrix]

there are 2 main ideas here. what we are differentiating, and what we are differentiating with respect to

[cloud saffron]

that seems to be what's unclear

to me

[brisk matrix]

what is

[cloud saffron]

that conept

concet

concept

usually it's just d/dx

but sometimes it's d/dy

unclear as to when to use each one

[brisk matrix]

the bottom variable is what we are differentiating with respect to

[cloud saffron]

ok

[brisk matrix]

im sure you have seen this before

we derive a constant, we get 0

[cloud saffron]

yes

[brisk matrix]

but if we change the notation slightly to

$\dv{c} (c)$

[warm shale]

maximo

[brisk matrix]

then we should get 1, because c is now our variable

that we are differentiating with respect to

[cloud saffron]

ok

now that makes sense

[brisk matrix]

now

$\dv{x} f(x)$

[warm shale]

maximo

[brisk matrix]

this is the derivative of a function with respect to x

a shorthand for this could be $\dv{x} f$

[warm shale]

maximo

[brisk matrix]

this doesn't mean f is a constant

and it also doesn't mean the derivative is 1

the derivative is simply written as $\dv{f}{x}$ or $f'(x)$

[warm shale]

maximo

[brisk matrix]

there are other notations but these are the most common

[cloud saffron]

ok let me process that

one second going to take some notes

[brisk matrix]

the reason i wrote it as f'(x) and df/dx is because we lack information about what f(x) is
if f(x) = x, then df/dx = 1
if f(x) = x^2, then df/dx = 2x
if f(x) = ln(x), then df/dx = 1/x

but in general, $\dv{x} f = \dv{f}{x} = f'(x)$

[warm shale]

maximo

[cloud saffron]

ok yes that's leibniz vs newtonian notation

same thign

[brisk matrix]

yes

the point is

as long as we don't know what f(x) is

we just write f', df/dx, or something else to represent the derivative

because we don't know what it is yet

[cloud saffron]

ok

[brisk matrix]

similarly

$\dv{x} y = y'$

[warm shale]

maximo

[brisk matrix]

because y is implied to be a function of x

and we do not know what y is, so we just leave the derivative as y', or dy/dx

note that y is a function of x just like f, or g, or h, or whatever

it's just a function of x, nothing more

[cloud saffron]

ok

yes

can you give me a short example?

[brisk matrix]

yes

$\dv{x}(y^2)$

[warm shale]

maximo

[brisk matrix]

now again

imagine this as

$\dv{x}(f(x)^2)$

[warm shale]

maximo

[brisk matrix]

which, if we let g(x) = x^2

can be written as

$\dv{x} g(f(x))$

[warm shale]

maximo

[brisk matrix]

so we have a composition of functions

[cloud saffron]

one second

f(x)^2

[brisk matrix]

yes

[cloud saffron]

wouldn't it be g(f(x)^2)?

[brisk matrix]

no

g(x) = x^2

so g(f) = f^2

or g(f(x)) = f(x)^2

[cloud saffron]

ok

yes

g(f(x))

a composition of functions

[brisk matrix]

then the derivative becomes

$\dv{x} g(f(x)) = g'(f(x)) \cdot f'(x)$

[warm shale]

maximo

[brisk matrix]

g(x) = x^2

g'(x) = 2x

so this becomes

2f(x) * f'(x)

this should hopefully make sense as it's just chain rule

nothing new has happened

[cloud saffron]

yes

what is f prime of x

[brisk matrix]

we don't know

that's the point

we don't know what f(x) is so we just adopt the notation f'(x) and keep moving forward

if we return to our use of y instead

we get

$\dv{x} y^2 = 2y\cdot y'$

[warm shale]

maximo

[brisk matrix]

again, nothing has actually changed

we're just going back to using y's instead of f(x)'s

[cloud saffron]

yes

[brisk matrix]

so the sole reason we need the chain rule is that y is a function of x

so y(x) could be x^2

it could be ln(sqrt(x))

we don't know

so we just call it y and y' and move on

[cloud saffron]

ok

my whole misunderstanding was the fact that y^2 was a composite function

why is y not just a variable again?

[brisk matrix]

y is a function of x

[cloud saffron]

ah yes

because of the notation

[brisk matrix]

y takes the form y(x) = something of x

[cloud saffron]

d/dx

[brisk matrix]

this notation means we're differentiating with respect to x

since y is a function of x

[cloud saffron]

ok

[brisk matrix]

we need to consider what happens when we differentiate y

another way you can visualize this

is

$\dv{x} y^2 = \dv{x} (y\cdot y)$

[warm shale]

maximo

[brisk matrix]

we can use product rule here

this will yield $(\dv{x}y)\cdot y + y\cdot (\dv{x}y)$

[warm shale]

maximo

[brisk matrix]

but what is d/dx of y?

it's just dy/dx

because we don't know what y is, but we do know it's a function of x

[cloud saffron]

ok

one unrelated question

how can i improve my logical and quantitative thinking? i make mistakes in calculations almost always because i can't apply the mathematical logic. my thought process never follows this logical flow and that is why i make mistakes.

i see numbers as symbols and the thought process and logic

[brisk matrix]

you need to understand definitions well in order to apply them

if you understand what a derivative is, this should follow nicely

i would review the definition of a derivative, what dy/dx means notationally, and what differentiation is in general (geometrically and analytically)

[cloud saffron]

ok

maths just often seems so abstract and out of context

when you look at practical application it makes more sense

[brisk matrix]

calculus is one of the most applicable aspects of math

you can find tons of examples for every single calculus concept that will be thrown at you

[cloud saffron]

but often times we aren't shown the application and it becomes rote memorization of arbitrary formulas and theorems

i don't know how to see past that

ok

[brisk matrix]

if you don't understand the theory, then there is no other way than to memorize

the only way you can apply without memorizing is to understand what you're doing

[cloud saffron]

i think i should review the theory first and maybe the logic will follow

[brisk matrix]

which doesn't come along unless you understand the definitions you are being given

that's the only way it'll follow

you cannot gain intuition without understanding what we're talking about
if i told you to "asoifdnaosuefaovn" the "faosindfaosiefn" you'd have no idea what to do nor could you understand it intuitively

[cloud saffron]

true

[brisk matrix]

similarly, if i tell you to "differentiate" an "arbitrary function of x", you won't be able to unless you understand what that entails

[cloud saffron]

ok

differentiation is just finding the slope on a curve

[brisk matrix]

that's the geometric aspect yes

[cloud saffron]

but we don't have a given point

so if we have an arbitrary function

how can we differentiate?

[brisk matrix]

well that's the thing

if we are just told

"f is a function of x" that is "f(x) is a function of x"

then the derivative of f is just... f' or f'(x)

that's all we can say

[cloud saffron]

true

[brisk matrix]

heck we don't even know if the derivative exists

but we assume it does for our purposes

[cloud saffron]

ok

[brisk matrix]

so we just give it a name

now

i want you to also understand why we are giving the derivative a name in the first place

most of these implicit differentiation problems take the form of

1. you are given an equation of x's and y's

2. you differentiate both sides

3. you isolate dy/dx to find the general derivative of y

which i think on its own is pretty cool, given that you can find the derivative of the curve, without actually having the curve itself as a function (that may be hard to process if you dont understand the vocabulary)

[cloud saffron]

having the curve itself as a function

[brisk matrix]

that is

having the curve given by y(x) = some function of x

but for example

a circle is given by x^2 + y^2 = r^2

and we can't really write the whole circle as a single function without losing some info

[brisk matrix]

this is probably a good practice problem for implicit differentiation btw

[cloud saffron]

i've seen the example before in lecture i think

it's a classic example

[brisk matrix]

yup

we basically get that a circle has the derivative $\dv{y}{x} = -\frac{x}{y}$

[warm shale]

maximo

[brisk matrix]

at all points

which is pretty cool

[cloud saffron]

are able to find that because we divide the circle into a top half and a bottom half?

[brisk matrix]

well no, we get that by implicit differentiation

[cloud saffron]

ok

[brisk matrix]

the left side differentiates into 2x + 2y * y'

and the right side to 0

it simplifies to y' = -x/y

[cloud saffron]

ok

[brisk matrix]

so

in short

we're giving y' a name

so we can isolate it and find the slope of the curve

so now, we have $x^4 + x^2y^2 + y^3 = 5$

[warm shale]

maximo

[brisk matrix]

let's try implicit differentiation on this

[cloud saffron]

ok

[brisk matrix]

the first term is clear, 4x^3

[cloud saffron]

yes

[brisk matrix]

let's skip the middle term

let's take care of the y^3

we want to compute $\dv{x} (y^3)$

[warm shale]

maximo

[brisk matrix]

now, y is not just a variable

[cloud saffron]

not 3y^2?

ah yes

forgot

as we're given that notation

y is a function of x

[brisk matrix]

well y is implied to be a function of x

but it doesnt matter

[cloud saffron]

but one question

[brisk matrix]

yes

[cloud saffron]

why were we able to differentiate the first term using the power rule and not here with y^3?

ah i see

because one is x

and one is y

[brisk matrix]

yes

we are always differentiating with respect to x

but here we have y(x)^3

so we need to do the chain rule

[cloud saffron]

how is it y(x)^3?\

[brisk matrix]

y is implicitly a function of x

so we basically have y= y(x)

[cloud saffron]

ok

[brisk matrix]

so let's apply the chain rule

first we take the derivative of y^3 as you did to get 3y^2

but we multiply times the derivative of y

now recall what i said earlier:

1. we don't know what y is, so dy/dx is also unknown to us
2. we still want to give it the name y' or dy/dx, so we can isolate it later

so the derivative becomes $3y^2 \cdot \dv{x}(y) = 3y^2\cdot y'$

[warm shale]

maximo

[cloud saffron]

ok

[brisk matrix]

the derivative of 5 is clearly 0 since it is constant

so the last thing left to derive is x^2y^2

can you try to do it and i'll check your answer?

[cloud saffron]

ok

i'm working through it now

one second

am i applying the chain rule correctly so far?

[brisk matrix]

no

first notice that this is the product of 2 functions

y^2 and x^2

[cloud saffron]

yes

[brisk matrix]

so we want to do product rule first

[cloud saffron]

ok

one second

applying product rule here first?

[brisk matrix]

no

well

you applied product rule first which is good

but again

the derivative of y^2

is 2y * y'

and you can't add those 2 together

[cloud saffron]

ok yes

that's right

one second

[brisk matrix]

yes

now put it all together

[cloud saffron]

can i combine like terms this way?

[brisk matrix]

yes

well

no

you can't do that

[cloud saffron]

ok. just keep them separate then? what have i done wrong in combining like terms there?

[brisk matrix]

remember you can only combine terms if their powers for their variables are the same

one has y^2, but the other doesn't

[cloud saffron]

as in y prime?

ok. so we can't combine any like terms there

we're left with that second to the last step then

[brisk matrix]

3y^2 y' + x^2 y' =/= 3y^2x^2 * 2y'

tthat doesn't work

[cloud saffron]

we leave it this way then?

[brisk matrix]

yes

now put all the terms without y' on the right side

[cloud saffron]

now isolate

and solve right?

[brisk matrix]

yes

[cloud saffron]

how to combine the y primes here?

[brisk matrix]

factor

the y'

[cloud saffron]

out

ok

made a mistake one second

i've made a mistake somewhere but not sure where

[brisk matrix]

what makes you say that

[cloud saffron]

it doesn't match with the answer in the back of the textbook

[brisk matrix]

can i see the answer?

[cloud saffron]

the numerator is just factoring out but my denominator is not right

[brisk matrix]

the derivative for $x^2y^2$ is $2x^2yy' + 2xy^2$

[warm shale]

maximo

[brisk matrix]

you're missing a y

[cloud saffron]

which line?

[brisk matrix]

2xy^2 + x^2y'

it should say

$2xy^2 + 2x^2yy'$

[warm shale]

maximo

[cloud saffron]

i'm missing 2 as the coefficient and a y

[brisk matrix]

yes

[cloud saffron]

where did i go wrong here

let's see

f(x) is x^2

and g prime of x is y prime

not sure where i went wrong in applying the chain rule

[brisk matrix]

yo want the derivative of y^2

so we do 2y * d/dx (y)

= 2y * y'

so we get x^2 * 2y * y'

[cloud saffron]

why are we breaking up y^2 there?

so we do 2y * d/dx (y)

[brisk matrix]

breaking up what

yes

[cloud saffron]

from that

[brisk matrix]

well we're finding $\dv{x} x^2y^2$

[warm shale]

maximo

[brisk matrix]

so we do product rule

[cloud saffron]

in our case it would be x^2 * g orime of x

[brisk matrix]

hold on

[cloud saffron]

g prime of x is y^2 here

[brisk matrix]

let's take it step by step

$\dv{x}(x^2y^2) = (\dv{x}x^2)y^2 + (\dv{x}y^2)x^2$

[warm shale]

maximo

[cloud saffron]

product rule

[brisk matrix]

yes

then $\dv{x}x^2$ is just $2x$

[warm shale]

maximo

[cloud saffron]

yes

[brisk matrix]

and $\dv{x}y^2 =2yy'$

[warm shale]

maximo

[cloud saffron]

don't see that hold on

we differentiate using the power rule

then add y prime onto the end?

[brisk matrix]

we need the chain rule here

remember

y is a function of x

so it's derivative isn't just 1

[cloud saffron]

but it looks like we've differentiated y^2 using the power rule and then added y prime onto the end as we're taking the derivative of y with respect to x

[brisk matrix]

that's exactly what we're doing

it's the chain rule

[cloud saffron]

it looks like two different steps

[brisk matrix]

let's do it slowly

$\dv{x}(y^2) = \dv{x}(y(x)^2)$

[warm shale]

maximo

[cloud saffron]

why is y^2 a composite function again?

[brisk matrix]

because y is a function of x

[cloud saffron]

ok yes

[brisk matrix]

so we have f(y(x)) where f(x) = x^2

[cloud saffron]

that's right

ok

[brisk matrix]

then we get f'(y(x)) * y'(x)

which is just 2y * y'

so 2yy'

[cloud saffron]

one second

chain rule is f prime * g(x) * g prime(x)

g prime of x

[brisk matrix]

f'(g(x)) * g'(x)

[cloud saffron]

is g prime of x

just g prime * (our x value)?

[brisk matrix]

no

[cloud saffron]

given f(x)

[brisk matrix]

for us, g(x) = y(x)

so

derivative of g(x) = derivative of y(x)

derivative of y(x) is just y'(x)

or y'

[cloud saffron]

isn't our g(x) y^2 here?

[brisk matrix]

mo

no

f(x) = x^2

g(x) = y

we've done this quite a few times if you read above

[cloud saffron]

ok one second

and $\dv{x}y^2 =2yy'$

[warm shale]

maple_wheats

[cloud saffron]

i see this as take the derivative of y^2 then add y prime

to the end

as it's the chain rule

but don't see the logic

i just accept it to be true

we're taking the derivative of y^2 with respect to x

so we get

2y * y prime

by first applying the power rule

then adding y prime to the end

that's how i'm interpreting it

[brisk matrix]

that’s basically all there is to it

we have

$\dv{x} f(g(x))$

[warm shale]

maximo

[brisk matrix]

where f(x) = x^2

and g(x) is arbitrary

the derivative becomes

f’(g(x)) * g’(x)

f’(x) = 2x

so that becomes

2g(x) * g’(x)

now g is arbitrary so we could also just call it y

we get

2y(x) * y’(x)

[cloud saffron]

so let's use our original values now

we apply the formula above as the chain rule

and we get

y^2 * y * y prime

d/dx of the above

[brisk matrix]

where did that come from

[cloud saffron]

no

[brisk matrix]

y^2 * y * y’

[cloud saffron]

g(x) is y^2

[brisk matrix]

what is this tho

[cloud saffron]

a mistake

trying to apply the chain rule

we have x^2*y^2 as our original term

so f(x) = x^2

[brisk matrix]

no

hold on

before we apply the chain rule we should apply the product rule

[brisk matrix]

to get this

now we can apply the chain rule