# What is wrong with this reasoning in with respect to working with differential of z

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 '18 at 2:18

## 3 Answers

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. – warrior_monk Sep 23 '18 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. – warrior_monk Sep 25 '18 at 11:50