Help Appreciated

li@mo

Can anyone tell me what this is:

d/dy (2xy -xSin(xy))

Its the -x before the Sin thats causing the problem.

Is this done so using the Product Rule?

Thanks

li@mo

li@mo wrote: »

Can anyone tell me what this is:

d/dy (2xy -xSin(xy))

Its the -x before the Sin thats causing the problem.

Is this done so using the Product Rule?

Thanks

Implicit differentiation? Yah, you use the product rule to differentiate the -x and sin(xy) together. But you'll have to use the chain rule as well for the sin(xy), while using the product rule for the (xy), while differentiating it implicitly.

[(-x)(cos(xy)([dx/dy]y+x)]+[(-dx/dy)sin(xy)]

You'll have to use the product rule for the 2xy as well.

I think what I've done is right. But I could be wrong. Wait until someone else responds to make sure. And perhaps there's an easier way of doing it.

AlanOB

Looks good to me.

li@mo

Thanks guys......Ill give it a go

I was trying another question there in the meantime on stationary points and im having problems with that aswell. I might post it here later.

li@mo

li@mo wrote: »

Thanks guys......Ill give it a go

I was trying another question there in the meantime on stationary points and im having problems with that aswell. I might post it here later.

Yah post whatever problems you have and I'm sure someone here will be able to help you.

timbrophy

li@mo wrote: »

Can anyone tell me what this is:

d/dy (2xy -xSin(xy))

Its the -x before the Sin thats causing the problem.

Is this done so using the Product Rule?

Thanks

When you are getting d/dy the x behaves as if it were a constant. Your question will be answered as follows:
d/dy(2xy -xSin(xy)) = d/dy(2xy) - x d/dy(Sin(xy))
= 2x -x Cos(xy) x
=2x - x^2 Cos (xy)

timbrophy

timbrophy wrote: »

When you are getting d/dy the x behaves as if it were a constant. Your question will be answered as follows:
d/dy(2xy -xSin(xy)) = d/dy(2xy) - x d/dy(Sin(xy))
= 2x -x Cos(xy) x
=2x - x^2 Cos (xy)

You'd only do that if x and y aren't related. You just got the partial derivative with respect to y. Maybe x is a funtion of y, then you'd implicitly differentiate it. The OP should come back and clarify.

timbrophy

I agree, but I suspect that x and y are independent.

LeixlipRed

I'd suspect it too. Very rarely would you be asked to differentiate wrt to y if it's a function in one variable. Partial differentiation is what the OP is looking for I'd say. They might come back and clarify.

LeixlipRed

LeixlipRed wrote: »

I'd suspect it too. Very rarely would you be asked to differentiate wrt to y if it's a function in one variable. Partial differentiation is what the OP is looking for I'd say. They might come back and clarify.

Yah ye are probably right. I suppose it's a bit odd to say x is a function of y, it's almost always the other way around.

li@mo

Sorry guys only coming bak now.

The question is actually on Taylors Series

Use Taylors series to expand to 2nd order terms, the function:

f(x,y) = xy^2 + Cos(xy) about the point (1,Pi/2)

The bit Im stuck on is 2nd order differentiation.

li@mo

li@mo wrote: »

Sorry guys only coming bak now.

The question is actually on Taylors Series



The bit Im stuck on is 2nd order differentiation.

Ya you take the partial derivative with respect to y, just as timbrophy explained. Completely ignore what I wrote down.

li@mo

-JammyDodger- wrote: »

Ya you take the partial derivative with respect to y, just as timbrophy explained. Completely ignore what I wrote down.

Thats great, thanks.....I was thinking it was a bit easier than that.

li@mo

li@mo wrote: »

The bit Im stuck on is 2nd order differentiation.

Sorry, I didn't read this bit before I posted my previous comment.

I haven't done much with the Taylor series, but to get the second order derivative, just find the second partial derivative of the function with respect to y, then sub into the formula (I think, but again, I could be wrong).

li@mo

-JammyDodger- wrote: »

Sorry, I didn't read this bit before I posted my previous comment.

I haven't done much with the Taylor series, but to get the second order derivative, just find the second partial derivative of the function with respect to y, then sub into the formula (I think, but again, I could be wrong).

ya.........i have to get the 2nd order derivative with respect to x and y.
i reckon if i get the y one then x should be straight forward.

I dont have it in front of me so itll be tomorro before i look at it.

thanks again

LeixlipRed

With partial differentiation you differentiate wrt to one variable and hold all other variables constant. For example if you have F(x,y,z)=2xyz then the derivative wrt to x is 2yz, wrt to y is 2xz and wrt to z is 2xy. You just use all your normal rules from differentiation of functions with one variable but remembering that we treat the other variables as if they were constants (ie numbers!).

li@mo

Cheers guys. I have it sorted now. Its the differentiation thats letting me down. I need to brush up on it.

Another question i was doing:

Find the stationary values of the function:

f(x,y) = x^3 - 6xy + y^3

I started as follows

df/fx = 2x^2 - 6y = 0

and

df/dy = -6x + 3y^2= 0

By solving these I get:

y=x^2/2

x=0 and x= 4 root(18)


This just doesnt seem right to me.
Can anyone clarify.

Thanks

When you solve for y, 3x^2 - 6y = 0 you get y = (x^2)/2; and when you solve y, 3y^2 -6x = 0 you get y = sqrt(2x).

Equating the two terms gives us sqrt(2x) = (x^2)/2; then solving this for x gives us that x is equal to 2. Subbing 2 into either of the above two equations gives us a value of 2 for y. So the stationary point would be (2,2).

(There's no guarantee of what I've done above, as I've actually never done multi-variable calculus:pac: But, I think it looks right!)

li@mo

sound right to me now. Thanks

Im going through a past paper here that I failed so I may have lots of problems....ill try them all myself first but i hope you guys can help me a bit if i get stuck.

LeixlipRed

By my calculations there are two stationary points. One is the origin (easy to see that) and the other is a bit of a nasty one. (2,2) doesn't satisfy both equations.

LeixlipRed

LeixlipRed wrote: »

By my calculations there are two stationary points. One is the origin (easy to see that) and the other is a bit of a nasty one. (2,2) doesn't satisfy both equations.

If the two partial derivatives are 3x^2 - 6y and 3y^2 - 6x, then does (2,2) not satisfy it along with (0,0)?

I think the OP made a mistake with the partial derivative with respect to x, they've written 2x^2 - 6y, where I think it should be 3x^2 - 6y.

LeixlipRed

Is it not 2x^2 in the first equation?

LeixlipRed

Oh sorry, didn't read your post fully. If they made that mistake then those would be the two points. That would leave you needing to solve x^3=8 which gives you x=2. Otherwise you need to solve x^3=18 which is nasty.

LeixlipRed

LeixlipRed wrote: »

Oh sorry, didn't read your post fully. If they made that mistake then those would be the two points. That would leave you needing to solve x^3=8 which gives you x=2. Otherwise you need to solve x^3=18 which is nasty.

No you probably read it fully, I just went back to edit it to point out the mistake!

timbrophy

(0,0) and (2,2) are the two points.

The OP had a typo where df/dx should be 3x^2 - 6y.


Edit: Sorry. I am late with this

li@mo

Ya I made a mistake. I took the question down wrong....tried and failed to complete it.

Put it up here with the same mistake.

Then when I was looking at Jammy Dodgers solution I copped the mistake........which was probably why I couldnt solve it in the first place.

I should have pointed that out here earlier. Sorry but thanks for the help