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/sci/ - Science & Math

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1970578 No.1970578 [Reply] [Original]

hey there /sci/
sorry to bug you with a math problem, but if anyone would solve this quick issue I'm having

derive with respect to x
xy^3

>> No.1970582

is it y^3 + 3y^2x?

>> No.1970583

Derive? Do you mean differentiate?

>> No.1970586

3xy^2 + c

>> No.1970581

y^3

>> No.1970590

>>1970583
Yeah, sorry.

>> No.1970591

I think that you make y^3 as a constant and derive x which results in y^3

>> No.1970595

Assuming y is a function of x, y^3 + x3y^2y'.

>> No.1970596

So if you're finding the derivative with respect to x, that means you treat y like a constant (assuming this is a partial derivative).
The derivative of x is 1

Therefore the answer would be y^3
like >>1970581 said

>> No.1970603

>>1970591
Yes, assuming y is a constant. If not, then you have:

y = xy^3
dy/dx = 3xy^2dy/dx + y^3
(1-3xy^2)dy/dx = y^3
dy/dx = (y^3)/(1-3xy^2)

I think that's probably what OP is doing. Although his stupidity makes it hard to discern his meaning.

>> No.1970610

Chain rule into quotient rule you fucking tards.

>> No.1970613

>>1970603
But then you could just divide both sides by y, and get:

1 = xy^2
y = sqrt(1/x)

>> No.1970625
File: 1 KB, 94x36, xy^3.gif [View same] [iqdb] [saucenao] [google]
1970625

you got a lot of answers from people... give a try to a computer...
http://www.wolframalpha.com/input/?i=x(y^3)+differenciate

and I was right

>> No.1970633

>>1970625
That's because there's a lot of different ways to interpret OPs question.

>> No.1970637

>>1970603
this you fucking inbreed

implicit differentiation

>> No.1970643
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1970643

>>1970625
> mfw that's the partial derivative, and everyone fucking knows how to do that

>> No.1970648

>>1970625
I'm supposed to find the slope of a tangent line given an equation using implicit differentiation. The only part of it I'm stuck on is xy^3.
Using y^3 isn't working but I'll give it another try.

>> No.1970651

>>1970637
Are you calling me or the OP an inbred? Just making sure I'm not being stupid.

>> No.1970653

he said derive not derivative you r tards

>> No.1970654

>>1970648
The answer is right there: (>>1970603)

What more do you want? Seriously...

>> No.1970669

>>1970651
calling OP the inbreed. sorry for confusion. your math was correct

>> No.1970711

none of these answers have a y' in it, which is what I need.
fuck.

>> No.1970719

>>1970711
3y^2y'

>> No.1970721

Hey guys sorry to interrupt the thread I need a huge favor.

Derive the following for me: x
Thanks a bunch /sci.

>> No.1970728

>>1970719
derp meant 3y^2dy/dx

>> No.1970733

>>1970721
d/dx(x^2)/2

>> No.1970738

Ok I've made like 4 posts in this thread because it's so fucking stupid.

Derive with respect to x means it's is a partial derivative. Look it up.

The first post is right.

/thread

>> No.1970764

>>1970711
god you are a fucking moron. seriously

>> No.1970767

>>1970711
are you really that dumb?

>> No.1970800

give us a fucking equation OP. such as xy^3 = k. then tell us to differentiate with respect to x.well then we would do
y^3 = k/x
y = (k/x)^(1/3)
y' = 1/3*k/x^(-2/3) = 1/3*k*x^(2/3)
k is a constant, whatever is on the other side of the equation.

>> No.1970814

>>1970800
sorry the answer should all be to the 2/3 power, and you flip the 1/3k to 3k.

>> No.1970996

xy^3 + 3y - 13.9 = 0
then I have a point.
just need to solve so that y' = .....