I know implicit differentiation. Just playing around with the differential of $z$ I did something like this, which I am sure is wrong somewhere. I want to know where I am wrong. Given a function $z = f(x,y)$, its differential is

\begin{align} \delta z &= \frac{\partial z}{\partial x} \cdot \delta x + \frac{\partial z}{\partial y} \cdot \delta y \end{align}

Now dividing by $\delta x$ throughout and taking the limit $\delta x$, $\delta y$, $\delta z$ tending to zero, I get something like

\begin{align} \frac{\partial z}{\partial x} &= \frac{\partial z}{\partial x} \cdot 1 + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial x} \end{align}

Cancelling the $\frac{\partial z}{\partial x}$ term, I get $\frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial x} = 0 $, which is not an identity.

My question is, where am I wrong?

  • If we were allowed to treat $\partial x$ and $dx$ (which you note $\delta x$) as interchangeable (and cancelable), then we'd have a more basic problem: in the original equation we could cancel $\delta x$ and also $\delta y$ and we'd get $\delta z = 2 \delta z$... – leonbloy Sep 23 at 2:18
up vote 3 down vote accepted

It's the difference between the partial derivative $\frac{\partial z}{\partial x}$ and the total derivative $\frac{\mathrm{d} z}{\mathrm{d} x}$. The total derivative accounts for potential dependency of $x$ and $y$ on each other. In our case, we would have $$\frac{\mathrm{d} z}{\mathrm{d} x} = \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} \frac{\mathrm{d} y}{\mathrm{d} x}.$$ Note that this is precisely the (single-variable) derivative of a two variable function $z$, with $x$ and a function $y(x)$ substituted in. Rearranging this gives us the differential of $z$ as before.

$$\mathrm{d} z = \frac{\partial z}{\partial x}\mathrm{d} x + \frac{\partial z}{\partial y} \mathrm{d} y.$$

Your mistake was treating $\mathrm{d} z$ and $\partial z$ as interchangeable, which they aren't.

  • Thanks,it makes sense now. – Abhishek_04 Sep 23 at 3:04

You are dividing $$\begin{align} \delta z = \frac{\partial z}{\partial x} * \delta x + \frac{\partial z}{\partial y} * \delta y \end{align}$$

by $ \delta x$ to get $$ \begin{align} \frac{\partial z}{\partial x} = \frac{\partial z}{\partial x} * 1 + \frac{\partial z}{\partial y} * \frac{\partial y}{\partial x} \end{align}$$

How did you get $$\frac{\partial z}{\partial x}$$ on the $LHS$ ?

Then you cancelled two different things from both sides.

Somehow you are confused between partial derivatives and total derivative.

In the right-hand side you don't have $z=z(x,y)$, but the composition $\tilde{z}(x)= \tilde{z}(x,y(x))$. It is another different function. The chain rule gives $$\frac{{\rm d}\tilde{z}}{{\rm d}x}(x) = \frac{\partial z}{\partial x}(x,y(x))+\frac{\partial z}{\partial y}(x,y(x))\frac{{\rm d}y}{{\rm d}x}(x). $$You were a victim of the abuse of notation $\tilde{z}\equiv z$. In fact, I am commiting one more abuse myself: using $y$ both for a variable of the function $z$ and to denote a function of $x$ (but I suppose you understand that no one wants to keep writing $\tilde{z}(x)=z(x,\tilde{y}(x))$, etc.).

  • Thanks. Understand now. – Abhishek_04 Sep 25 at 11:50

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