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Mathematics 12 Online
OpenStudy (anonymous):

Cab someone help me understand.... Given: z=z(x,y) find... dz/dy partial if: x^4+3xy^2+5z=sin(z)

OpenStudy (anonymous):

alright for this treat everything as a constant except for Y

OpenStudy (anonymous):

and if you are differentiating with z you have to put dz/dy and solve for that specific value. Or if you are in multivariable calculus there is a formula which is -Fy/Fz would give your answer

OpenStudy (anonymous):

this is basically like implicit differentiation but with 2 variables and if you remember for one variable when you differentiated y you had to put dy/dx so same case applies

OpenStudy (anonymous):

@hamza_b23 I love your straight explanation :)

OpenStudy (anonymous):

thank you

OpenStudy (anonymous):

do i use the formula then or do i differentiate with respect to y and when i get to z i put dz/dy?? sorry implicit differentiate i never understood...

OpenStudy (anonymous):

use the formula if you like you would get same answer

OpenStudy (anonymous):

so just differentiate with respect to z and y and take the negative quotient?

OpenStudy (anonymous):

and do i set the equation to 0 or leave it as is?

OpenStudy (anonymous):

if you are finding dz/dy the implicit for that is -Fy/Fz

OpenStudy (anonymous):

so take the partial for y over the partial for z with negative as the coefficient

OpenStudy (anonymous):

@drifta808 just follow hamza's instruction one step at the time!

OpenStudy (anonymous):

( x^4 ) ' = ....

OpenStudy (anonymous):

formula looks much easier lol

OpenStudy (anonymous):

If it's much easier, why don't you post your solution here?

OpenStudy (anonymous):

-(6xy)/5-cos(z)?

OpenStudy (anonymous):

Good job, Very close :)

OpenStudy (anonymous):

do i leave sin(z) on other side of equal sign? making it -(6xy)/5 = cos(z)

OpenStudy (anonymous):

cant be right...

OpenStudy (anonymous):

I'm unsure if you mistype or not, just a tiny little bit different? z' = - 6xy/ ( 5z - cosz )

OpenStudy (anonymous):

why is it 5z? and not 5? dont u take the derivative?

OpenStudy (anonymous):

Since you take dz: d (5z) = 5z dz/dy

OpenStudy (anonymous):

that is what i dont understand...

OpenStudy (anonymous):

That's how implicit work out!

OpenStudy (anonymous):

Pretending take implicit dy/dx with z as y

OpenStudy (anonymous):

z now is function of y, just as y is function of x!

OpenStudy (anonymous):

ok thanks, gonna have to watch some videos on implicit differentiation.. forget everything about it during multivariable calc

OpenStudy (anonymous):

Mee, too! I tend to forget so easily :(

OpenStudy (anonymous):

ok, i still dont understand why dz/dy of 5z = 5z(dz/dy) i get the (dz/dy) part but doesnt the chain rule make it just 5 * (dz/dy)

OpenStudy (anonymous):

its suppose to be 5*dz/dy

OpenStudy (anonymous):

so the answer i put was right? its -6xy/(5-cos(z))?

OpenStudy (anonymous):

6xy+5(dz/dy)-cos(z)(dz/dy)=0 dz/dy(5-cos(z))=-6xy dz/dy=-6xy/(5-cos(z)) ??

OpenStudy (anonymous):

that the right way to solve this?

OpenStudy (anonymous):

yeah your right bro

OpenStudy (anonymous):

ah ok thanks alot man... other guy made me really confused there but now i know how to solve these.. thanks!

OpenStudy (anonymous):

can u tell me if im doing another problem right?

OpenStudy (anonymous):

its all good use the formula its really simple alright if you see dz/dy then its -Fy/Fz

OpenStudy (anonymous):

yeah sure

OpenStudy (anonymous):

yea i like the formula more too but its nice to know both and this next one is supposed to be a total differential...

OpenStudy (anonymous):

f(x1,x2...,xn)=sin(x1+2x2+3x3+...+nxn)

OpenStudy (anonymous):

i got cos(x1)+2cos(x2)+...+ncos(xn)

OpenStudy (anonymous):

is that all it is?

OpenStudy (anonymous):

so what are you doing differentiating that?

OpenStudy (anonymous):

it says find the total differential...

OpenStudy (anonymous):

O ok one min

OpenStudy (anonymous):

and yes bro your absolutely right, but make sure you put in your deltax1 delta x2 ..... delta xn as well

OpenStudy (anonymous):

did you get that? so your answer would just be cos(x1)deltax1+2cos(x2)deltax2+...+ncos(xn)deltaxn

OpenStudy (anonymous):

delta is the triangle notation

OpenStudy (anonymous):

oh sorry got distracted with another problem but ok that makes sense thanks!!! ready to destroy this test now haha

OpenStudy (anonymous):

lmao goodluck do me a favourite and give me a medal in every post :P im trying to beat my friend lmao

OpenStudy (anonymous):

haha for sure will do!

OpenStudy (anonymous):

awesome goodluck

OpenStudy (anonymous):

thanks... u 2

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