Functions equal implies equal partial derivative.

Question

Suppose there are two surfaces described by the functions $f(x,y)$ and $g(x,y)$. The functions are equal along some line $y(x)$ i.e. $f(x,y(x))=g(x,y(x))$. I realise it's naive to next write $f(x,y(x))=g(x,y(x)) \Rightarrow \partial f/\partial y = \partial g / \partial y$. My question is why is it naive to say this?

Answer 1

The partial derivatives of $f$ and $g$ are their slopes as we move in the $y$ direction holding $x$ constant, not their slopes in the direction of a tangent to the curve $y(x)$, so your "naive" claim is not, in general, true.

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A counterexample to your "naive" claim:

The functions $f$ and $g$ given by $f(x,y)=x^2y$ and $g(x,y)=xy^2$ are equal along the line $y(x)=x$, but

$$f_y(x,y)=x^2\neq 2xy= g_y(x,y)$$

unless $x=0$ or $y=x/2$.

Note that it is not even true that $f_y(x,y(x))=g_y(x,y(x))$.

What is true is that the directional derivatives of the two functions in the direction $(1,1)$ (along the line $y(x)=x$) are equal:

$$(f_x(x,x),f_y(x,x))\cdot (1,1)=(2x^2,x^2)\cdot (1,1)=3x^2$$

and

$$ (g_x(x,x),g_y(x,x))\cdot (1,1)=(x^2,2x^2)\cdot (1,1)=3x^2$$

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What we can say, given $f(x,y(x))=g(x,y(x))$ and $y$ differentiable is that:

$$f_x(x,y(x))+f_y(x,y(x))y'(x)=g_x(x,y(x))+g_y(x,y(x))y'(x).$$

From this we can deduce that:

1. If $y'(x)=0$, then $f_x(x,y(x))=g_x(x,y(x))$.
2. If $y'(x)\neq 0$ and $f_x(x,y(x))=g_x(x,y(x))$ then $f_y(x,y(x))=g_y(x,y(x))$

In (1.) the curve is a horizontal line ($x$ is changing, $y$ is constant), so the equality of the functions along the line implies they have the same slope in the $x$-direction, i.e. their partial derivatives with respect to $x$ are the same along the line $y(x)$.

Answer 2

A partial derivative doesn't take into account the possible dependency between the arguments. You compute it by considering the distinguished argument as the variable, while the others are constant.

So there is no reason that

$$\frac{\partial f(x,y)}{\partial y}\color{red}=\frac{\partial g(x,y)}{\partial y}$$ or

$$\frac{\partial f(x,y)}{\partial x}\color{red}=\frac{\partial g(x,y)}{\partial x}$$ unless by coincidence.

On the other hand, if you consider the functional relation between $x$ and $y$, you can write

$$\frac{df(x,y(x))}{dx}=\frac{\partial f(x,y(x))}{\partial x}+\frac{\partial f(x,y(x))}{\partial y}\frac{dy(x)}{dx} \\=\frac{dg(x,y(x))}{dx}=\frac{\partial g(x,y(x))}{\partial x}+\frac{\partial g(x,y(x))}{\partial y}\frac{dy(x)}{dx}.$$