Partial Derivate

[exotic cargo]

i need to find the partial derivate of x please help

[glad ibex]

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[twin finch]

@exotic cargo are y and z functions of x ?

[exotic cargo]

yea but they are 0 since i only need to find the derivate of x

i gotta get rid of the squareroot

[twin finch]

no no you can't take their derivative as 0

[twin finch]

yee we will get to that shortly

[twin finch]

if y is function of x

and we have say

y^2

take it like chain rule

[rose parrot]

treat everything except x as a constant

[exotic cargo]

yea

[twin finch]

they are functions of x tho ?

[rose parrot]

but its with respect to x

not y or z

[exotic cargo]

everything that has not a x it's 0

[twin finch]

if we have function of x say f(x)

we differentiate (f(x))^2 answer is gonna be 2f(x)f'(x) not 0 ?

[exotic cargo]

1/(1-x^2-y^2-z^2)^1/2

[rose parrot]

a partial derivative is only differentiating with respect to a variable

[exotic cargo]

this is the next step

then

[twin finch]

that would be if y and z were not functions of x right ?

quotient rule

[rose parrot]

no

its a power rule and then chain rule

but treat y and z as a constant

[exotic cargo]

i gotta use d/dx [u^n] = nu^n-1 * derivate of u

[twin finch]

its 1/ unless we make power -

both work

[rose parrot]

yeah but quotient rule is pointless

it makes it harder for you

[exotic cargo]

so what should i do

[rose parrot]

we have

[twin finch]

yee you are right

[rose parrot]

$(1-x^2-y^2-z^2)^{-1/2}$

[quaint torrent]

ビジヨン

[exotic cargo]

where the -1/2 comes from?

[rose parrot]

because you have it in the denominator

you move it to the numerator and it becomes a negative exponent

[exotic cargo]

down to top u mean?

[rose parrot]

its the same thing as

$1\over (1-x^2-y^2-z^2)^{1/2}$

[quaint torrent]

ビジヨン

[exotic cargo]

so i put in on top

[rose parrot]

yes

[exotic cargo]

oh

[rose parrot]

now use power rule

then chain rule

but while using chain rule everything will become 0

except the x

[exotic cargo]

$1/a^m'

[rose parrot]

you have to differentiate x

[exotic cargo]

how u do that?

[rose parrot]

$x$

[quaint torrent]

ビジヨン

[exotic cargo]

$1/overa^m'$

[quaint torrent]

Corr
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[exotic cargo]

$1/over(a^m')$

[quaint torrent]

Corr
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[rose parrot]

$1\over a^{m’}$

[quaint torrent]

ビジヨン

[exotic cargo]

yea like that xd

[rose parrot]

ok

[exotic cargo]

that's the one i gotta use?

[rose parrot]

just use power rule

$(1-x^2-y^2-z^2)^{-1/2}$

[quaint torrent]

ビジヨン

[rose parrot]

on this

then chain rule

[exotic cargo]

nx^n-1?

-1/2-1/2 will be?

[rose parrot]

yes

$n$ $x^{n-1}$

[quaint torrent]

ビジヨン

[exotic cargo]

$1/2(1-x^2-y^2-z^2)^1-/2$

[quaint torrent]

Corr

[exotic cargo]

$1/over(2)(1-x^2-y^2-z^2)^1-1/over(2)$

[quaint torrent]

Corr

[rose parrot]

thats the wrong answer for power rule

it’s supposed to be

$x\over (1-x^2-y^2-z^2)^{3\over 2}$

[quaint torrent]

ビジヨン

[rose parrot]

Do you see how I got the answer?

[exotic cargo]

hmm i got confused

[rose parrot]

so we use the power rule

we subtract 1 from -1/2

giving us -3/2 in the exponent

[exotic cargo]

oh

[rose parrot]

and 1/2(1-…-…-…)^3/2

[exotic cargo]

then we put it back on down?

[rose parrot]

yeah

now we differentiate the inside with respect to x

[rose parrot]

which if we solve, it will give us

[exotic cargo]

and whats left on the top i derivate it?

[rose parrot]

Ill send a picture tomorrow. Im tired

but the answer is there so you can give it a shot and understand the process

[exotic cargo]

okeyyy, thankkss!!

[rose parrot]

See you later.