1. ## 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.

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)

2. Hello, Parton-Bill!

You are pretty sloppy with your Algebra . . .

If . $z \:=\:e^x\cos(xy)$

find the partials for $x$ and $y$, and second order partial for $y.$

$\frac{\partial z}{\partial x} \:=\:e^x\cdot [-\sin(xy)]\cdot y + e^x\cdot\cos(xy) \;=\; e^x\bigg[\cos(xy) - y\sin(xy)\bigg]$

$\frac{\partial z}{\partial y} \;=\;e^x\cdot[-\sin(xy)]\cdot x \;=\;-xe^x\sin(xy)$

$\frac{\partial^2z}{]\partial y^2} \;=\;-xe^x\cdot\cos(xy)\cdot x \;=\;-x^2e^x\cos(xy)$

3. 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