#stupid question

102 messages · Page 1 of 1 (latest)

worthy cape
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The derivative of x^2 is 2x right?
So the derivative of 1^2 is 2 ?

hot mortarBOT
worthy cape
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Idk

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It should either be 0 or 2

grand sparrow
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No it should be 0 because 1^2 is a constant

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You should remember that is done with respect to a specific variable

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So when youre differentiating with x then youre looking for terms containing x

worthy cape
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Do you know partial derivative ? @grand sparrow

grand sparrow
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Terms not containing x is a constant respect to x and becomes zero

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What im explaining also applies to partial derivation yes

worthy cape
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For partial derivation i have a question but i dont know how to say the fancy math words in english@grand sparrow

grand sparrow
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Hmm

worthy cape
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Im gonna try to say it anyway

grand sparrow
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Ask the question anyways

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Yea

worthy cape
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So if gradient vector is (1 and -1 )
That mean i should replace x with 1 and y with -1 ?

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For a partial derivation of 3y-2x^(3)y

grand sparrow
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Is this a directional derivative question?

worthy cape
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To find the gradient vector of a function we need to use partial derivation right?

grand sparrow
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Yes

worthy cape
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So we should use the nabla thing

grand sparrow
worthy cape
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Yeah i think

grand sparrow
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Ok then

worthy cape
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So like i just need to replace x with 1 and y with -1?

grand sparrow
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gradient function of the given scalar field 3y-2(x^3)y is (its partial derivative wrt x, its partial derivative wrt y)

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no

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so you have to find the case where the x partial derivative of the given function is 1 and the y partial derivative of the given function is -1

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so it's going to be in the form of a system of equations

worthy cape
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How?

grand sparrow
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do you have a specific part you have trouble with

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or should i guide you through the solution

worthy cape
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So for the example i did partial derivation for y ( i only derivate y) and replace y with -1 and i found the answer is this not how you do it?

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∂f(x,y)/∂y = 3-2x^3

grand sparrow
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uhh no?

worthy cape
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Sorry i meant i replaced x with 1

grand sparrow
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no no

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so

worthy cape
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So i get 3-2

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And the result was 1

grand sparrow
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the partial derivative of the function wrt y should be equal to -1

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so 3-2x^3 = -1

worthy cape
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Oh

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Wait

grand sparrow
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and the partial derivative of the function wrt x should be equal to 1

worthy cape
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Wdym by wrt?

grand sparrow
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so -6(x^2)y = 1

grand sparrow
worthy cape
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Oooooh

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Right

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Idk why my wrong method gave the right result

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@grand sparrow what did i do wrong?

IMG_20221227_144837_693.jpg
agile zinc
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then you can *-1 on both sides

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2x^3=4

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divide by two

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x^3=2

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and then take to the power of one third

worthy cape
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Oh right

agile zinc
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x=2^(1/3)

worthy cape
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True

worthy cape
agile zinc
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so that they are both positive

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i guess you can divide by -2 too (to skip the step after)

worthy cape
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Wait

agile zinc
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what do you mean you are supposed to find 1, the solution is 2^(1/3)

grand sparrow
worthy cape
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Whats that?

grand sparrow
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the solution

worthy cape
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Of the gradient vector at the coordonate (1; -1) ?

grand sparrow
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no

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when the gradient vector is (1, -1)

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that's how i interpreted your question anyway

worthy cape
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Oh wait is that not the same thing

grand sparrow
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if you want to find the gradient vector at the coordinate (1, -1) then as you said you can just put it in the gradient function

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which is (-6(x^2)y, 3-2x^3)

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you said "So if gradient vector is (1 and -1 ) " so i interpreted your question as the former

worthy cape
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Yeah sorry about that

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So for (-6(x^2)y i transform it into -6(x^2) *-1 ?@grand sparrow

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Wait

grand sparrow
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yeah and also put 1 in x

worthy cape
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-6(1^2) *-1

grand sparrow
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yep that's right

worthy cape
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Which is 6

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Okay thank you i get it

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Thanks for helping me and spending ur time

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Appreciate it a lot

grand sparrow
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np\

grand sparrow
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.solved

quick heath
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it is definately zero because the power rule holds true when f(x) = x^n not some constant to the power of n

viral kiln
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op's question was solved, but since the channel hasn't been closed and people might still be reading it, here's a fun "proof" of 0=1 you can use to sanity check questions like if d/dx ( 1^2 ) equals 2

let x = 1
take derivative of both sides: d/dx (x) = 1, d/dx (1) = 0
and since x = 1, and therefore also d/dx (x) = d/dx (1), we conclude that 1 = 0

;3

snow willow
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if you treat 1 as a variable

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then

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d/d1 1^2 = 2*1

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but that's not 2 lol

snow escarp
snow willow
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$\dv{1} 1^2$

hexed scaffoldBOT
snow willow
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oh that is cursed

sand beacon
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No because 1 ^2 =1 and if f(X)=1
f'(X)=0 derivative is 0 because the gradient at all points of a line y=1 is 0 so it doesn't make sense pls someone correct me if I'm wrong

tawny tusk
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.solved