Thread: could some body please check these for me

Hi guys, Please could someone check the following for me, I have no confidence at all with calculus. have bought three books to day to finally get to grips with it, but these are due before then.

Many thanks in advance.

If: z = e^x . Cos (xy) Find the partials for x and x, and second order partial for y.

I have dz/dx = e^x . y - sin (xy) dz/dy = e^x . x - sin (xy)

And

d/dy * dz/dy = d^2 z/ dy^2 = e^x . x^2 . cos (xy)

Hello, Parton-Bill!

You are pretty sloppy with your Algebra . . .

If .

find the partials for and , and second order partial for

Much appreciated, thank you for you time.

Find an Expression for DY/DX (I use D as i will use d for partial derivatives)

When: x^3 + y^3 - 2x^2 y = 0

i) Assign;

z = x^3 + y^3 - 2x^2 y = 0 (eqn 1)

we also have Ez = dz/dx . Ex + dz/dy . Ey ( Where E = Delta/Change)

ii) Divide Both Sides By Ex

which then equates to ;

Ez/Ex = dz/dx + dz/dy . Ey / Ex (eqn 2)

and when Ex approaches zero eqn1 becomes;

DZ/DX = dz/dx + dz/dy . DY/DX (eqn 3)

iii) we can now find the partial derivatives;

dz/dx = 3x^2 - 4xy

dz/dy = 3y^2 - 2x^2


and since z = zero we can now substitue these back into eqn 3 to get;

0 = 3x^2 - 4xy + (3y^2 - 2x^2) . DY/DX

- 3x^2 + 4xy = (3y^2 - 2x^2) . DY/DX

DY/DX = (-3x^2 + 4xy) / (3y^2 - 2x^2)


Think thats not far off, sorry for changing the script and making it confursing.

Thanks again