Some key things to remember about partial derivatives are:
- You need to have a function of one or more variables.
- You need to be very clear about what that function is.
- You can only take partial derivatives of that function with respect to each of the variables it is a function of.
So for your Example 1, $z = xa + x$, if what you mean by this to define $z$
as a function of two variables,
$$z = f(x, a) = xa + x,$$
then $\frac{\partial z}{\partial x} = a + 1$ and
$\frac{dz}{dx} = a + 1 + x\frac{da}{dx},$ as you surmised,
though you could also have gotten that last result by considering $a$ as a
function of $x$ and applying the Chain Rule.
But when we write something like
$y = ax^2 + bx + c,$ and we say explicitly that $a$, $b$, and $c$ are
(possibly arbitrary) constants, $y$ is really only a function of one variable:
$$y = g(x) = ax^2 + bx + c.$$
Sure, you can say that $\frac{\partial y}{\partial x}$ is what happens
when you vary $x$ while holding $a$, $b$, and $c$ constant, but that's
about as meaningful as saying you vary $x$ while holding the number $3$ constant.
I suppose technically $\frac{\partial y}{\partial x}$
is defined even if $y$ is a single-variable function of $x$,
but it would then just be $\frac{dy}{dx}$ (the ordinary derivative),
and I can't remember seeing such a thing ever written as a partial derivative.
It would not make it possible to do anything you cannot do with
the ordinary derivative, and it might confuse people (who might try to
guess what other variables $y$ is a function of).
The previous paragraph implies that the answer to your Example 3 is "yes."
It also hints at why I almost wrote "a function of two or more variables"
as part of the first requirement for using partial derivatives.
Technically I think you only need a function of one or more variables,
but you should want a function of at least two variables before you
think about taking partial derivatives.
For Example 2, where we have $x^2 + y^2 = 1$, it is not obvious
what the function is that we would be taking partial derivatives of.
Either $x$ or $y$ could be a function of the other.
(The function would be defined only over a limited domain,
and would produce only some of the points that satisfy the equation, but
it can still be useful to do some analysis under those conditions.)
If you write something besides the equation to make it clear that
(say) $y$ is a function of $x$, giving a sufficiently clear idea which
of the possible functions of $x$ you mean, then I think technically you
could write $\frac{\partial y}{\partial x}$, and you might even find that
$\frac{\partial y}{\partial x} = 2x$, but again this is a lot of trouble
and confusion to get a result you could get simply by using
ordinary derivatives.
On the other hand, suppose we say that
$$h(x,y) = x^2 + y^2 - 1,$$
and we are interested in the points that satisfy $x^2 + y^2 = 1$,
that is, where $h(x,y) = 0$.
Now we have a function of multiple variables, so we can do interesting
things with partial derivatives,
such as compute $\frac{\partial h}{\partial x}$ and
$\frac{\partial h}{\partial y}$ and perhaps use these to look for trajectories
in the $x,y$ plane along which $h$ is constant.
OK, we don't really need partial derivatives to figure out that
those trajectories will run along circular arcs, but we could have
some other two-variable function where the answer is not so obvious.
In each of these problems, the idea is to simplify the difference quotient so that the denominator does not tend to zero as $h\to 0$.
For instance, let $f(x,y) = x^2y$. This is not your problem, but once you understand this problem, you will know how to do your problems. Then
\begin{align*}
f_x(x,y) &= \lim_{h\to 0} \frac{f(x+h,y) - f(x,y)}{h}
\\&= \lim_{h\to 0} \frac{(x+h)^2y - x^2y}{h}
\\&= \lim_{h\to 0} \frac{(x^2+2xh+h^2)y - x^2y}{h}
\\&= \lim_{h\to 0} \frac{x^2y+2xhy+h^2y - x^2y}{h}
\\&= \lim_{h\to 0} \frac{2xhy+h^2y}{h}
\\&= \lim_{h\to 0} (2xy+hy) = 2xy
\end{align*}
Best Answer
This is not the intention of the exercise. They do not want you to indentify the graph. To explain how they want you to do it, I will do 5(a).
It asks to find $f_x(1,2)$, i.e. how $f$ changes as you move in the $x$ direction from the point $(1,2)$. If you go in the positive $x$ direction from the pink spot which represents $(1,2)$, then you can see that the $z$ co-ordinate (i.e. $f$ value) increases. So the sign of $f_x$ is positive at this point.
Similarly you can do 5(b), 6(a) and 6(b).
When it asks to find $f_{xx}(-1,2)$, this means "the rate at which $f_x$ is changing as you advance in the $x$ direction at this point". At that point, $f_x$ is negative, since the graph is going downwards as you advance in the positive $x$ direction. But $f_x$ is increasing, since the "negative slope" is becoming less steep, like going from gradient of $-2$ to $-1$. So $f_{xx}$ is positive.
Similarly you can do 7(b).
When the two letters are different, you need to find "the rate at which $f_x$ changes as you move in the positive $y$ direction". This, combined with the method for question 7, is how to do 8(a) and 8(b).