can somebody please check the following for me. many thanks

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

2. $x^3 + y^3 -2x^2y = 0$.

$\frac{d}{dx}(x^3 + y^3 - 2x^2y) = \frac{d}{dx}(0)$

$\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) -2 \frac{d}{dx}(x^2y) = 0$

$3x^2 + \frac{d}{dy}(y^3)\,\frac{dy}{dx} -2 \left(x^2\,\frac{dy}{dx} + 2xy\right) = 0$

$3x^2 + 3y^2\,\frac{dy}{dx} - 2x^2\,\frac{dy}{dx} - 4xy = 0$

$\left(3y^2 - 2x^2\right)\frac{dy}{dx} = 4xy - 3x^2$

$\frac{dy}{dx} = \frac{4xy - 3x^2}{3y^2 - 2x^2}$.

3. There are NO "partial derivatives" here. y is a function of the single variable x.

4. there has to be, its part of an assignment so the question would not be designed to catch us out. thanks though

5. Originally Posted by Parton-Bill
there has to be, its part of an assignment so the question would not be designed to catch us out. thanks though
And why would you assume that there has to be a partial derivative? The question asks you to find $\frac{dy}{dx}$, which you can easily do by implicit differentiation. Also note that you got the same answer.

If you're not sure, ask your instructor - guessing usually doesn't work, and to anyone else trying to solve this it would not seem that there this has anything to do with multi-variable functions, so until you get that solved out we can't help you.

6. to be clearer, the question states:

Use partial Differentiation to determine an expression for dy/dx.

so although it can be solved without partial derivatives, my task is to do so.

sorry, should have been clearer but thanks for the advice anyway

7. Originally Posted by Parton-Bill
to be clearer, the question states:

Use partial Differentiation to determine an expression for dy/dx.

so although it can be solved without partial derivatives, my task is to do so.

sorry, should have been clearer but thanks for the advice anyway
"Implicit differentiation" of, say f(x,y)= 0, can be written as " $\frac{\partial f}{\partial x}+ \frac{\partial f}{\partial y}\frac{dy}{dx}= 0$" so that $\frac{dy}{dx}= \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$
but since implicit differentiation is usually introduced well before partial derivatives, I would consider that a confusing notation.

8. in honesty the entire assignment has been confusing, it was even worse as an exam. alot comes down to what the british term as education, we are taught to jump through hoops, not use best judgment, I suppose judgment comes later when i'm designing building by numbers lol