What is the Simplification of the Last Term in Differentiation?

[pyroknife]

Hey guys. I have a question regarding the differential operator d.

Say we have an equation d(Z+X*Y^2)=0
If we want to differentiate the expression in the parenthesis, are these the steps to follow?
d(Z+X*Y^2)=0
dZ+d(X*Y^2)=0

Apply product rule to the second term:
dZ+Y^2*dX+X*dY^2=0

Here is where I get confused. To simply the 3rd term (X*dY^2), is the simplification this:
2Y*X*dY or 2*X*dY?

 

[Mark44]

Hey guys. I have a question regarding the differential operator d.

Say we have an equation d(Z+X*Y^2)=0
If we want to differentiate the expression in the parenthesis, are these the steps to follow?
d(Z+X*Y^2)=0
dZ+d(X*Y^2)=0

Sort of. Assuming that f is a function of x, y, and z, then the total differential df involves the three partials.
In other words, $$ df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz$$

Apply product rule to the second term:
dZ+Y^2*dX+X*dY^2=0

Here is where I get confused. To simply the 3rd term (X*dY^2), is the simplification this:
2Y*X*dY or 2*X*dY?

You should get this:
$$\frac{\partial (xy^2)}{\partial x} dx + \frac{\partial (xy^2)}{\partial y} dy $$
For each partial, treat the other variable as if it were a constant. Is that clear?

 

[pyroknife]

Got it. Thank you.

I don't know how to write out in the format you did, but for that last term partial(x*y^2)/partialy*dy,
I was a little confused on how that simplifies.
We hold x constant for that term, so this gives:
x*partial(y^2)/partial_y*dy
Does this give
x*2*y*partial(y)/partial_y*dy
or
x*2*partial(y)/partial_y*dy?

 

[Mark44]

Got it. Thank you.

I don't know how to write out in the format you did, but for that last term partial(x*y^2)/partialy*dy,
I was a little confused on how that simplifies.
We hold x constant for that term, so this gives:
x*partial(y^2)/partial_y*dy
Does this give
x*2*y*partial(y)/partial_y*dy
or
x*2*partial(y)/partial_y*dy?

This -- x*2*y*partial(y)/partial_y*dy -- which simplifies to 2xy dy. The partial of y with respect to y is just 1.

I wrote my previous reply using LaTeX, which isn't too difficult. It looks like this:
\frac{\partial f}{\partial x}
Put a pair of $ symbols at front and back, and it renders like this:
$$\frac{\partial f}{\partial x}$$