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I'm having trouble convincing myself of the validity of the following set of steps in differentiation of a multi-variable function under a constraint:

Suppose $f(x,y) = 0$

The above constraint implicitly makes $x$ dependent on $y$. I can do the following operation \begin{align} \frac{df(x,y)}{dx} = 0 \end{align} Since the derivative is not partial, $y$ is allowed to change with changing $x$ and the constraint $f(x,y) = 0$ can still be satisfied. I know that taking partial derivative wouldn't be legal since that would make $y$ constant and the constraint will no longer be satisfied.

How do I prove this formally? That

\begin{align} \frac{{\partial f(x,y)}}{{\partial x}} \end{align}

is invalid given that $f(x,y) = 0$.

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Well, that's two different concepts: the partial derivative and the total derivative.

Assuming $x$ and $y$ are the independent variables in the function $f(x,y)$, we can define its partial derivatives $$\frac{\partial f}{\partial x}=\frac{\partial f(x,y)}{\partial x} \quad \text{and} \quad \frac{\partial f}{\partial y}=\frac{\partial f(x,y)}{\partial y}.$$

Assuming $y$ is also defined implicitly as a function of $x$, we can say that we have two functions: a function of two variables $f(x,y)$, where we can still treat $y$ is one of the independent variables; and a function of a single variable $x$ defined as $g(x)=f(x,y(x))$. I think what you're trying to find is the derivative of this new function: $$\frac{dg}{dx}=\frac{df}{dx}=\frac{df(x,y(x))}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\cdot\frac{dy}{dx}.$$

Note that in this context, both notations $\displaystyle \frac{\partial f}{\partial x}$ and $\displaystyle \frac{df}{dx}$ make sense but mean different things.

Moreover, when an equation such as $f(x,y)=0$ defines $y$ as an implicit function of $x$, this is the basis for the implicit differentiation method for finding the derivative of $y$ with respect to $x$: $$f(x,y)=0 \implies \frac{df}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\cdot\frac{dy}{dx}=0 \implies \frac{dy}{dx}=-\frac{\partial f/\partial x}{\partial f/\partial y}.$$

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  • $\begingroup$ There's no definition involved in $\frac{\partial f}{\partial x}=\frac{\partial f(x,y)}{\partial x}$ -- that's just shorthand notation on the LHS. $\endgroup$ – Noldorin Aug 15 '17 at 23:07
  • $\begingroup$ @Noldorin: Yes, I know. I didn't mean to say that I was defining them there -- I wasn't. I only said that we could define the following partial derivatives, which can be denoted like this, and provided two notations for each (for convenience of future reference). $\endgroup$ – zipirovich Aug 15 '17 at 23:14
  • $\begingroup$ Ah okay. Slightly ambiguous wording, perhaps, but no worries now that you've clarified it. Each mathematician has their own style, of course. $\endgroup$ – Noldorin Aug 15 '17 at 23:15
  • $\begingroup$ @Noldorin: Sure, no problem. Thanks for forcing me to clarify this! :-) $\endgroup$ – zipirovich Aug 16 '17 at 1:32
  • $\begingroup$ Sure, hopefully will help the OP, or someone else. :-) $\endgroup$ – Noldorin Aug 16 '17 at 1:36
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I've always thought it from the definition of the derivative:

$$ \frac{df}{dx} = \lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

Multivariable versions are just slightly different:

$$ \frac{\partial f}{\partial x} = \lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x,y)-f(x,y)}{\Delta x} $$

and

$$ \frac{\partial f}{\partial y} = \lim_{\Delta y \rightarrow 0}\frac{f(x,y + \Delta y)-f(x,y)}{\Delta y} $$

Obviously (x,y) is a parameter in all the versions... Important property is obviously the linearity: $$ L(a+b) = L(a)+L(b) $$ $$ L(c a) = c L(a)$$, where you can choose $L=\frac{\partial f}{\partial x}$ or $L=\frac{\partial f}{\partial y}$

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