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Let's say I have function $z=f(x,y)=x^2+xy+y^2$

Please verify if my understanding of partial derivative is correct..I have put it in my words below:

I understand that a partial derivative with respect to $x$ gives me equation of slope if $y$ is kept constant. When we keep $y$ constant, we take a plane perpendicular to $y$ axis, curve $z$ will intersect that plane and that intersection will be a line or a curve. The partial derivative with respect to $x$ is $2x+y$ and therefore that is the equation of tangent at any point on that line. That tangent line will be parallel to $x$ axis.

Now if we take partial derivative of (partial derivative of $x$) with respect to $y$ what do I get? In this case it will be derivative of $2x+y$ and it will come out to be $1$. Does it has any meaning? Also what is the total derivative in this case? what does total derivative give? In case of $z$ curve if I want to find coordinates of point that has the maximum or minimum value then how could we calculate it?

thanks

I tried wiki search but couldn't understand the concepts. I also looked at a few coursera courses but I felt that they explain calculations than explaining the concept.

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  • $\begingroup$ First, you can write formulas in $\LaTeX$ using dollar symbols. Second, there is a mistake in the partial derivative of $2x+y$. $\endgroup$ Commented Oct 19, 2013 at 18:43
  • $\begingroup$ Will use dollar symbol going forward. Also made some changes to the post...I hope that corrects my mistake $\endgroup$ Commented Oct 19, 2013 at 18:47
  • $\begingroup$ updated my mistake now...sorry for the delay $\endgroup$ Commented Oct 19, 2013 at 21:41
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    $\begingroup$ A nice way to think about the derivative is $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$. (To approximate the change in $f$, just multiply the instantaneous rate of change of $f$ by the change in $x$.) For partial derivatives, we have $f(x+\Delta x,y) \approx f(x,y) + \frac{\partial f(x,y)}{\partial x} \Delta x$ and $f(x,y+\Delta y) \approx f(x,y) + \frac{\partial f(x,y)}{\partial y} \Delta y$. $\endgroup$ Commented Oct 19, 2013 at 21:53
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    $\begingroup$ I don't have any special intuition for that, beyond just $\frac{\partial f(x,y+\Delta y)}{\partial x} \approx \frac{\partial f(x,y)}{\partial x} + \frac{\partial^2 f(x,y)}{\partial y \partial x} \Delta y$. In other words $\frac{\partial^2 f(x,y)}{\partial y \partial x}$ is just the instantaneous rate of change of $\frac{\partial f}{\partial x}$ at $(x,y)$ in the $y$ direction. $\endgroup$ Commented Oct 19, 2013 at 22:39

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Maybe it's helpful to view at the partials from a more general point. Let $c$ be a differentiable curve defined on an open interval $I$ containing $0$ to $\mathbb R^2$ with $c(0)=p$ and $c'(0)=v$. The standard-example for such a curve is $c(t):=p+tv$. Let $f\colon G\mapsto\mathbb R$ defined on an open subset $G$ of $\mathbb R^2$. Let $p\in G$ and $v\in\mathbb R^2$. We define the derivative of $f$ in $p$ at $v$, namely $d_pf(v)$ via $$d_pf(v)=\frac{d}{dt}\Big|_{t=0}f\bigl(c(t)\bigr),$$ provided that limit exists. Then for a fixed $p$ the map $$ d_pf\colon\mathbb R^2\to \mathbb R\quad\text{is linear.}$$ Now we may want to find a matrix for that linear map in respect to the basis $\{(1,0)^T,(0,1)^T\}$ of $\mathbb R^2$ and $\{1\}$ of $\mathbb R$. All we need to do is to plug in the base vectors in $d_pf$. Well, to plug in $(1,0)^T$ we'll have to evaluate $$\frac{d}{dt}\Big|_{t=0}f\bigl(p+t(1,0)^T)\bigr),$$ i.e., treat $y$ as constant and differentiate $f$ in respect to $x$. We will henceforth call $$\frac{\partial f(p)}{\partial x}:=d_pf\bigl((1,0)^T\bigr)$$ the partial derivative of $f$ in $p$.

Thus the matrix of $d_pf$ is $$\left(\frac{\partial f(p)}{\partial x},\frac{\partial f(p)}{\partial y}\right).$$

Let's define $$\nabla_pf:=\left(\frac{\partial f(p)}{\partial x},\frac{\partial f(p)}{\partial y}\right)^T,$$ then presence of the standard dot product we finally gain $$d_pf(.)=\langle\nabla_pf(.),.\rangle.$$

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