4
$\begingroup$

I want to explain the question I have in mind with an example.

$ F(x,y,z) = 0 $ system of equations : Let $z$ define as a implicit function of $x$ and $y $ . And $F(x,y,z)=x^2-y+\ln(z)$

$(a)$ $\dfrac{\partial F}{\partial x}=F_x = 2x$

$(b)$ $\dfrac{\partial F}{\partial y}=F_y = -1$

$(c)$ $\dfrac{\partial F}{\partial z}=F_z = \dfrac{1}{z}$

$F(x,y,z)=0$

Let's take the partial derivative of both sides of the equation with respect to $y$ .

then $\dfrac{\partial}{\partial y}F(x,y,z)=\dfrac{\partial }{\partial y}(0)$ $=>$ $\dfrac{\partial F}{\partial y}=0$

$(b)$ owing to $\dfrac{\partial F}{\partial y}=-1$

as a result : $-1=0 $ is happening. Where am I doing wrong ? why did this happen ?

$\endgroup$
  • $\begingroup$ z = f(x,y) , then $ \partial_x z \ne 0$ and $\partial_y z\ne 0$ $\endgroup$ – Ajay Mishra Jul 18 at 0:19
4
$\begingroup$

What's going on is a very unfortunate abuse of notation. This is very common in differential calculus, and even more so in the context of implicit functions. It is only after you really understand what the function is (this includes knowing its domain and target space) vs where it is being evaluated that you'll understand what's going on here.

To start off, you're given a function $F: U \to \Bbb{R}$, where $U = \Bbb{R}^2 \times \Bbb{R}^+$ defined by \begin{align} F(x,y,z) = x^2 - y + \ln(z) \end{align} The statement "$F(x,y,z) = 0$ implicitly defines $z$ as a function of $x$ and $y$" is a convenient but somewhat imprecise way of saying the following more precise (and pedantic) statement:

There exists a function $\zeta: \Bbb{R}^2 \to \Bbb{R}$ such that for all $(x,y) \in \Bbb{R} \times \Bbb{R}$, we have that $F(x,y,\zeta(x,y)) = 0$.

I intentionally used $\zeta$ instead of $z$, because I think it is confusing (for a beginner atleast) to use the same letter $z$ in two different contexts, where they have different meanings. If you use $z$ to denote the "independent variable" in $F(x,y,z)$, and also as the "implicitly defined function" $z: \Bbb{R}^2 \to \Bbb{R}$, $(x,y) \mapsto z(x,y)$, then this is just a recipe for all sorts of confusion.

Now, the reason you are getting the apparent contradiction $-1 = \dfrac{\partial F}{\partial y}$ and $\dfrac{\partial F}{\partial y} = 0$ is because you're using the same letter $F$ to describe two different things. The correct and pedantic statement is:

For all $(x,y,z) \in U$, we have $(\partial_2F)_{(x,y,z)} = -1$.

(i.e we're taking the partial derivative of the function $F$ with respect to its second argument, and evaluating the function $\partial_2F$ at the particular point $(x,y,z) \in U$)

In slightly more common notation, we might say that for all $(x,y,z) \in U$, we have $\dfrac{\partial F}{\partial y} \bigg|_{(x,y,z)} = -1$.

Let us temporarily define a new function $g: \Bbb{R}^2 \to \Bbb{R}^3$ by the rule \begin{align} g(x,y) = (x,y,\zeta(x,y)) \end{align} Then, the first highlighted statement says that the composition $F \circ g = 0$ is the zero function; i.e for all $(x,y) \in \Bbb{R}^2$, $(F \circ g)(x,y) = 0$.

Later on, you make the statement:

\begin{align} \dfrac{\partial}{\partial y}F(x,y,z) = \dfrac{\partial}{\partial y}(0) = 0 \implies \dfrac{\partial F}{\partial y} = 0, \end{align}

which is riddled with all sorts of notational abuse (one of the several pitfalls of Leibniz notation). A more accurate statement is any of the following ones below:

  • $\partial_2(F \circ g) = 0 \, \,$ (this is an equality of functions)
  • For all $(x,y) \in \Bbb{R}^2$, $\partial_2(F \circ g)_{(x,y)} = 0 \, \,$ (this is pointwise version of the first statement, and is an equality of real numbers)
  • For all $(x,y) \in \Bbb{R}^2$, $\dfrac{\partial (F \circ g)}{\partial y}\bigg|_{(x,y)} = 0$

The first two are the most notationally precise way of stating things. The last one is probably more common/convenient, but as always Leibniz's notation should always be used with caution.


I hope now you notice that there is no contradiction at all. We have two completely different statements; the first is that for all $(x,y,z) \in U $, $(\partial_2F)_{(x,y,z)} = -1$. The second is that for all $(x,y) \in \Bbb{R}^2$, $\partial_2(F \circ g)_{(x,y)} = 0$.

If you insist on Leibniz's notation, we could say that $\dfrac{\partial F}{\partial y} = -1$, whereas $\dfrac{\partial (F \circ g)}{\partial y} = 0$.


Of course, now you can use the chain rule to get some relationships between the partial derivatives: for any $(x,y) \in \Bbb{R}^2$, \begin{align} 0 &= \partial_2(F \circ g)_{(x,y)} \\ &= (\partial_1F)_{g(x,y)} \cdot (\partial_2g_1)_{(x,y)} + (\partial_2F)_{g(x,y)} \cdot (\partial_2g_2)_{(x,y)} + (\partial_3F)_{g(x,y)} \cdot (\partial_2g_3)_{(x,y)} \\ &= (2x) \cdot (0) + (-1) \cdot (1) + (\partial_3F)_{(x,y, \zeta(x,y))} \cdot (\partial_2 \zeta)_{(x,y)} \\ &= -1 + \dfrac{1}{\zeta(x,y)} \cdot (\partial_2 \zeta)_{(x,y)}. \end{align}

BTW, in the above computation $g_i$ is the $i^{th}$ component function of $g$. So, $g_1(x,y) := x$, and $g_2(x,y) := y$, and $g_3(x,y) := \zeta(x,y)$.

What I have done above is the notationally precise way of doing things, because I was very careful to distinguish between which function I'm differentiating and where I'm evaluating the derivatives. The more convenient, and also more misleading (if you're just starting out) way of performing the same computation is: \begin{align} 0 &= \dfrac{\partial }{\partial y} \left( F(x,y,z(z,y))\right) \\ &= \dfrac{\partial F}{\partial x} \cdot \dfrac{\partial x}{\partial y} + \dfrac{\partial F}{\partial y} \cdot \dfrac{\partial y}{\partial y} + \dfrac{\partial F}{\partial z} \cdot \dfrac{\partial z}{\partial y} \\ &= (2x) \cdot (0) + (-1) \cdot (1) + \dfrac{1}{z} \cdot \dfrac{\partial z}{\partial y} \\ &= -1 + \dfrac{1}{z} \dfrac{\partial z}{\partial y} \end{align}

$\endgroup$
  • $\begingroup$ Thank you so much. You are a very good person . <3 $\endgroup$ – Tonyukuk Jul 18 at 20:51
3
$\begingroup$

As you said, $z$ is a imlpict function of $x$ and $y$, i.e. $z=z(x,y)$. So when you derivative both sides of $F(x,y,z)=0$ with respect to $y$, the left-hand side turns out to be $$\frac{\partial(x^2)}{\partial y}-\frac{\partial(y)}{\partial y}+\frac{\partial(\ln z)}{\partial y}=-1+\frac{\partial(\ln z)}{\partial z}\frac{\partial z}{\partial y}=-1+\frac1z\frac{\partial z}{\partial y}.$$

$\endgroup$
  • $\begingroup$ $F_y$ equal to what , $(-1)$ or $(-1+\frac1z\frac{\partial z}{\partial y}).$ ? $\endgroup$ – Tonyukuk Jul 18 at 0:24
  • $\begingroup$ @Tonyukuk If you regard $z$ as a function of $x,y$, then the latter one is right. $\endgroup$ – Feng Shao Jul 18 at 0:52
  • $\begingroup$ Thank you so much. You've made me very happy. i am thankful to you . $\endgroup$ – Tonyukuk Jul 18 at 1:02
  • $\begingroup$ @Tonyukuk You are welcome. $\endgroup$ – Feng Shao Jul 18 at 1:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.