differentiation ?

[neon marsh]

Anybody know how to solve this no. 4. Its not even my homework im just curios

[blazing kelp]

Have you ever heard of implicit differentiation?

know that $$\frac{dy}{dx} = \frac{dy}{du}*\frac{du}{dx}$$

[frigid ravine]

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[blazing kelp]

so if you took the derivative wrt x of both expressions in the problem

$$\frac{d}{dx}[2y^{2}-x^{3}] = \frac{d}{dx}[0]$$

[frigid ravine]

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[blazing kelp]

now try it

[blazing kelp]

note: you can rearrange this so

$$\frac{du}{dy}*\frac{dy}{dx} = \frac{du}{dx}$$

[frigid ravine]

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[neon marsh]

Hm

I got 4y-3x^2

But thatsnot answer in mark scheme

[blazing kelp]

$\frac{d}{dx}[2y^{2}] = \frac{d}{dy}[2y^{2}]*\frac{dy}{dx}$

[neon marsh]

Im so confused what this mean sorry

[frigid ravine]

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[blazing kelp]

you're very close with this, but compare that with the above

[neon marsh]

I mean

Firdt of all i dont understand what implicit differentaion actually is

[blazing kelp]

to be more clear d/dy [2y^2] = 4y

[neon marsh]

And i dont understand why i have to use it in this task but not the other onws

[blazing kelp]

what do you notice is the extra factor here

[neon marsh]

Times dy/dx

[blazing kelp]

exactly

so now retry your answer but with keeping dy/dx there

[neon marsh]

But

Isnt dy/dx what im finding

[blazing kelp]

yes, but why not treat it as a variable

just think of dy/dx as some variable that you're trying to solve for

[neon marsh]

Man

[blazing kelp]

and remember that what you're given is an equation, not just an expression

[neon marsh]

Yeah

But i cant treat it like quadratic equation

Since it has x2 and y

Not two x

[blazing kelp]

ok do you want me to write it out or do you want to think about it some more?

[neon marsh]

Idk i think its better if u write it out

Since my brain dodsnt undefstand it

[blazing kelp]

try evaluating these derivatives here

[neon marsh]

But actually i would appreciate rven more a reason why in all the ofher questions i could have found the differentiation of each of the factors and that was the end, but here i have to do other stuff

[blazing kelp]

what is the right-hand side equal to

d/dx [0]

[blazing kelp]

ok here's a method that might make more sense to you

try solving for y first in 2y^2 - x^3 = 0

[blazing kelp]

well you're still technically doing that

[blazing kelp]

as shown here

but you see that there's a part 2y^2 which is dependent on the variable x

if it were instead 2x^2 we could just do 4x

[blazing kelp]

but instead we must use

where u = 2y^2

but to give you a final solution for more direction

$4y\frac{dy}{dx}-3x^{2} = 0$

so, solving for dy/dx,

$\frac{dy}{dx} = \frac{3x^{2}}{4y}$

[frigid ravine]

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[robust fjord]

Is it not (3x^2)/4?

[neon marsh]

Wait

Ths is correct answer

Im gonna watch some video on this later ig

[blazing kelp]

one second I'll re-attack it in a bit

[neon marsh]

Thanks for the help man

Im still kinda lost

But this wasnt even my task so whatever

Just saw it and it looked easy so wantsd to do it

[blazing kelp]

yea I see my mistake but im in the middle of a timed test rn lmao

so one sec

[neon marsh]

But im confused

Oh ok

Gl

[blazing kelp]

(3x^2)/4y

[robust fjord]

Oh yea

Forgot the y

[blazing kelp]

alr im back

[blazing kelp]

ok so it looks like they solved it using what I was saying here:

they solved for y such that $y = \sqrt{\frac{x^{3}}{2}}$

[frigid ravine]

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[blazing kelp]

then, taking the derivative of y wrt x...
$\frac{dy}{dx} = \frac{1}{\sqrt{2}}\frac{d}{dx}[x^{3/2}]$

[frigid ravine]

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[blazing kelp]

and I think you have the ability to take it from there

it's the same thing you've been doing with an added step of first rearranging for y

but be wary that the assumption $y > 0$ is why this is true

[frigid ravine]

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