Partial derivatives with and without a composite function

Question

I have a simple partial differentiation calculation that illustrates my problem,

$ u = x $; $v = x + y + 1 $ and $ w = u + v $

so

$\frac{\partial w}{\partial u} = 1$; $\frac{\partial w}{\partial v} = 1$ and $\frac{\partial w}{\partial x} = 2$

Now if I formulate this in a different way, making $v$ a function of $u$,

$ u = x $; $v = u + y + 1 $ and $ w = u + v $

the partial derivative

$\frac{\partial w}{\partial u} = 1 + \frac{\partial v}{\partial u}= 2$

changes, but $\frac{\partial w}{\partial v}$ and $\frac{\partial w}{\partial x}$ remain the same.

My question is this: is it correct that $\frac{\partial w}{\partial u}$ changes from one formulation to the other, or have I made a mistake? The change seems counter-intuitive, because $v$ is still equal to $x+1$. On the other hand, $w$ is clearly twice as sensitive to $u$ in the second formulation.

The background to this question is that I have two ways of calculating a quantity. One uses composite functions, the other avoids them; both use the same set of input data. I get different values when I calculate partial derivatives of the final result wrt intermediate quantities that are equivalent but calculated differently (like $v$ in my simple case).

I believe that my simple example shows that different sensitivities do not necessarily mean that there is a mistake in the calculations. But perhaps I have missed something?

Answer 1

The answer is: no, the partial derivative $\frac{\partial w}{\partial u}$ should not change. The reason for the confusion is that the meaning of $\frac{\partial w}{\partial u}$ is ambiguous in the second form, because $u$ appears in the definition of $v$ as well as in the definition of $w$.

One interpretation of $\frac{\partial w}{\partial u}$, then, is the partial derivative of $w = u + v$ with respect to the symbol $u$ that appears explicitly in the equation, so \begin{equation} \frac{\partial w}{\partial u} = 1 \end{equation}

But another interpretation involves the chain rule for partial differentiation, because $v$ is a function of $u$, so

\begin{align} \frac{\partial w}{\partial u} &= 1 + \frac{\partial v}{\partial u} \\ &= 2 \end{align}

To resolve the ambiguity, the notation must be changed, as, for example, in the first form.

Answer 2

Both are same. And you don't have any mistake. You are replacing variables that is possible so it's fine.