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A lot of books basically say this on the subject of partial derivatives:

If $f(x,y)$ is a function, then $\frac{\partial}{\partial x}$ means: when $x$ increases by $1$ unit, then $f(x,y)$ changes by $\frac{\partial}{\partial x}$ units.

My question is about that $1$ unit thing. Isn't it supposed to be an increase of a very tiny $\epsilon$ amount that is very near of $0$ instead of being an increase of $1$ unit in the $x$ value?

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    I think they are just trying to say that $\frac{\partial f(\mathbf{x})}{\partial x}$ is the slope of the line tangent to the point $(\mathbf{x},f(\mathbf{x}))$ on the graph of $f$ along the direction parallel with the $x$-axis. It is clearly not meant to be a rigorous statement. You are right to think that said tangent line is a better approximation for $f$ the less the change in the $x$-direction. – 2017-02-14
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    You need to fire those books. – 2017-02-15

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Yes, you are correct. The derivative is defined to make $f(a,b)+\varepsilon \frac{\partial f}{\partial x}(a,b)$ a linear approximation to $f(a+\varepsilon,b)$ as $\varepsilon \to 0$, in that the difference between these two things is sublinear, i.e. in little-$o$ notation, $$ f(a+\varepsilon,b) = f(a,b)+\varepsilon \frac{\partial f}{\partial x}(a,b) + o(\varepsilon) \quad \text{as } \varepsilon \to 0. $$ Rearranging this gives the definition, $$ \frac{\partial f}{\partial x}(a,b) = \frac{f(a+\varepsilon,b) - f(a,b)}{\varepsilon} + o(1) \quad \text{as } \varepsilon \to 0. $$ $1$ is normally considered quite a large number compared to $\varepsilon$[citation needed], so you're right to query this usage; presumably the point they want to get across is that your first estimate of how much $f$ increases when you add one unit of $x$ is to add the partial derivative on, thinking about situations where $f$ has sufficiently low relative curvature and $1$ is a relatively small amount (I add one atom to my mole, I add one unit of currency onto my spending, ...) that this initial approximation is reasonable, rather than literally correct.

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    So, I am right thinking about it as the _"smallest increase imaginable"_ of $x$ will give us a change of $\delta f / \delta x$ in the derivative, instead of _"an increase of 1"_ of $x$? – 2017-02-15
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    In the limit, the change in the function is proportional to the change in $x$: a change by a sufficiently small $\varepsilon$ causes a change in the function of $\varepsilon \frac{\partial f}{\partial x}$. If the change in $x$ is very small, the change in the function is also very small, but if the gradient of $f$ is not zero, the proportion is finite (and given by the derivative). – 2017-02-15
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    So in this case $\delta f / \delta x$ is a **proportion** of change and not **_THE_ value** of change? (Hence this is why you multiply this proportion of change to the $\epsilon$ value). – 2017-02-15
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    Yes: if you instead change $x$ by $2\varepsilon$, the change in $f$ will be $2\varepsilon \frac{\partial f}{\partial x}$ (approximately), and so on. Think about the case of $f$ a linear function $ax+b$ to start with. Then $f(x+\varepsilon)-f(x) = a\varepsilon$, exactly. – 2017-02-15
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    And the more we are getting far from $\epsilon$ the more this won't be true anymore, right? – 2017-02-15
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    In general, yes. Obviously there are special cases where the function happens to end up near its tangent plane again, but in general, the linear approximation is only locally reasonable. – 2017-02-15
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This may be nothing more than normalizing a ratio to have a denominator of one. To illustrate, let‘s try some concrete units. If the position of an object at some point $t$ is measured in meters and time in seconds, then its instantaneous velocity—the derivative of position with respect to time—would be measured in some number of meters per one second. This doesn’t mean that if you let the object travel for a second from that instant in time, that it will go exactly the distance that this instantaneous velocity value would suggest.

That said, if instead of looking at the change in $f$ itself we look at the linear approximation to this change that is given by $f$’s partial derivatives, then if we move one unit in the $x_k$ direction, the approximation’s value does change by exactly ${\partial f\over \partial x_k}$ units. Of course, the error in this approximation might be quite large at this distance from the starting point.