math question

sebastianlieken

its part of a differentiation question i found that i dont know how to solve. can someone tell me how to solve it plz


v=e^(y.x^2)

wats dv???

Thomas_S_Hunterson

sebastianlieken wrote:

its part of a differentiation question i found that i dont know how to solve. can someone tell me how to solve it plz


v=e^(y.x^2)

wats dv???

Is y a constant?

sebastianlieken

yup y is a constant, as is x

Fremen

Which variable are you differentiating with respect to?
Could you post the whole question as it appears in the text?

Edit: if it's any help,

dv/dy = (x^2).(e^(y.(x^2))

dv/dx = 2xy(e^(y.(x^2))

Further edit:
I was using partial differentiation here, I think in the context of the LC, dan's reply below is more likely to be correct

ZorbaTehZ

They can't both be constants, cause then 'twould differentiate to zero.

sebastianlieken

oh sorry....duh...my mind is melting....

differentiating with respect to x

dan719

Okay assuming that x any are both variables use the formula 1 divided by the f(x) multiplied by the first derarative of the f(x) where here the f(x) is = y.x^2

And then just use implicit differentiation. (within product rule)

JC 2K3

If they're both constants, then dv = 0.

Are they both functions or is one a function and the other a constant?

ZorbaTehZ

dv/dx = 2yx.e^(y.x^2) methinks

tolosenc

v = e^(yx^2)
=> dv = (2yx).[e^(yx^2)].dx

Fremen

if y is nonconstant you need to put a dy/dx in there (as far as I remember)
it would look like

dv/dx = 2xy(dy/dx).(e^(y.(x^2)))

JC 2K3

^Thay's if y is a constant(lol, this thread is moving fast, talking to Zorba)

v=e^(y.x^2)
dv/dx = (e^(y.x^2))(2xy + (dy/dx)(x^2))

I think...

ZorbaTehZ

sebastianlieken wrote:

yup y is a constant, as is x

sebastianlieken wrote:

oh sorry....duh...my mind is melting....

differentiating with respect to x

Y is a constant. Differentiating with respect to x.

sebastianlieken

ok...i tell ya wat, the question is;

find all first and second partial derivitives of f(x,y) = x.e^(y.x^2)

i split it using prod rule to get u= x & V= e^(y.x^2)

....i should prob mention this isnt a leaving cert question, its college calculus. but the differentiation part is very much leaving cert. standard if i remember correctly, i was just never good with the complicated ones...

the answer, if it helps is; {x^5}{.e^[(y).(x^2)]}

ZorbaTehZ

(Firstly, my knowledge of partial derivatives is limited. . .)
I'm confused - question states it's a partial derivative, you want us to hold one of the variables constant then i assume? Or solve the prod rule problem taking both of them as variables?
Also that is the answer to the d/dx of the prod rule problem or what?

sebastianlieken

my knoledge is limited aswell, i was playin around with different methods, but im havin no luck either.

the answer is to the whole question

Fremen

My answer above used partial differentiation, but it only has the first order derivtives.
The trick is, when diff'ing with respect to x, pretend y is a constant. Rename it 'a' or something in your head
when diff'ing with respect to y, pretend x is a constant.

To get the second order derivatives, use the same trick, along with the product rule.

JC 2K3

Ah, so y is constant.

v=e^(y.x^2)
dv/dx = (e^(y.x^2))(2xy)

u = x
du/dx = 1

d/dx(uv) = (2yx^2)(e^(y.x^2)) + e^(y.x^2)
= (e^(y.x^2))( 1 + 2yx^2)

2nd derivative
(1 + 2yx^2) = w
(e^(y.x^2)) = z

dw/dx = 4yx
dz/dx = (e^(y.x^2))(2xy)

(1 + 2yx^2)(e^(y.x^2))(2xy) + (e^(y.x^2))(4yx)
= (e^(y.x^2))(2yx + 2(y^2)(x^3)) + (e^(y.x^2))(4yx)
=(e^(y.x^2))(6yx + 2(y^2)(x^3))

.........

Over our heads as LC students? Or have I made a gaping error....

ZorbaTehZ

Ok I did out all the partials and the only one that bares resemblance to that answer you gave is:

f[subset{yy}] = {2x^5}{.e^[(y).(x^2)]}

sebastianlieken

ZorbaTehZ wrote:

Ok I did out all the partials and the only one that bares resemblance to that answer you gave is:

f[subset{yy}] = {2x^5}{.e^[(y).(x^2)]}

ok obviously you know what you are doing, very close answer, but the factor of 2 throws it off a wee little bit. i got a written version of the answer off my friend, infortunatly she hasnt a notion of what its all about either. ill post it up

JC 2K3

w00t, I was right.

I didn't know you wanted it differentiated with respect to y also though.

{x^5}{.e^[(y).(x^2)]} wasn't the answer, it was one of the answers, the other being what I got: (e^(y.x^2))(6yx + 2(y^2)(x^3)).

JC 2K3

q = x.e^(y.x^2)
dq/dy = x^3(e^(y.x^2))

d^q/dy^2 = x^5(e^(y.x^2))

That part was pretty simple actually....

ZorbaTehZ

/cry
Took down the question wrong - had the power of 2 on the y - doh.

sebastianlieken

OH GOD!!!

I SEE IT.... i didnt realise there were 2 parts. grrr, lesson learnt... you can point and laugh now if youd like....:rolleyes:

JC 2K3

*points and laughs.

sebastianlieken

JC 2K3 wrote:

*points and laughs.

little interesting fact.... im doing aeronautical engineering...one day, your life will be in my hands:D .

'right, so the undercarriage of this plane were designing can support a total mass of 300 tons.... that means we can the plane weigh 300 tons...actually, that might be cutting it close, better take a little off those wings...'

mwa ha ha ha:D

JC 2K3

I'll be doing computer science next year. I'll be writing programs to replace incompetent fools like you!

I learned what partial differentiation was today. Yay!

Nehpets

So this won't be coming up then? <_< >_>

Fremen

CS is a good laugh.
Go get your "public static void main" on in the meantime

Seb, you forgot the 'x' multiplying into the e when you first asked the question

MathsManiac

Since this is nothing to do with the Leaving Cert, shouldn't it be somewhere else?

ZorbaTehZ

lol, bit late there mate. Question is finished, topic is over.