Derivative and Integral For Other Metrics Than the Absolute Valued Metric Perhaps this comes as too general of a question, but if we define a "limit" via a metric other than the absolute value metric, does the derivative of a function differ?  How does it differ?  What about the integral of a function?  What if we use, for example, the discrete metric d(x, y)=0 if x=y, d(x, y)=1 otherwise?
 A: If you use the discrete metric, then limits no longer make sense.
Recall the definition of limit of $f$ as $x\to a$: the limit of $f(x)$ as $x\to a$ is $L$ if and only if for every $\epsilon\gt 0$ there exists $\delta\gt 0$ such that if $0\lt d(x,a)\lt \delta$, then $d(f(x),L)\lt \epsilon$. For any $L$ and any $\epsilon$, selecting $\delta=\frac{1}{2}$ makes the definition vacuously true. So any $L$ is "the" limit of $f(x)$ as $x\to a$.
In general, limits don't make sense at isolated points; but in the discrete metric, every point is isolated: for every point $x$, there exists $\epsilon\gt 0$ such that the collection of all points $y$ with $d(x,y)\lt \epsilon$ is equal to $\{x\}$: just pick $\epsilon=\frac{1}{2}$. 
That also means that derivatives don't make sense, because they are defined as limits of a certain function (the difference quotient), but then the limit is at an isolated point.
If the metric you pick is strongly equivalent to the usual metric, then limits are the same. Recall that two metrics $d_1$ and $d_2$ are strongly equivalent if and only if there exist positive constants $\alpha$ and $\beta$ such that for all $x$ and $y$, 
$$\alpha d_2(x,y)\leq d_1(x,y) \leq \beta d_2(x,y).$$
In particular, $d_1(x,y)=0$ if and only if $d_2(x,y)=0$. 
To see that limits under $d_2$ equal limits under $d_1$, suppose that $\lim\limits_{x\to a}f(x) = L$ using $d_1$. That means that for every $\epsilon\gt 0$ there exists $\delta\gt 0$ such that if $0\lt d_1(x,a)\lt \delta$, then $d_1(f(x),L)\lt \epsilon$. To show this also holds for $d_2$, let $\varepsilon\gt 0$. Then there exists $\delta_1\gt 0$ such that if $0\lt d_1(x,y)\lt \delta_1$, then $d_1(f(x),L)\lt \frac{\varepsilon}{\alpha}$. Now let $\delta=\frac{1}{\beta}\delta_1$.
If $0\lt d_2(x,a)\lt \delta$, then $0\lt d_1(x,a)\leq \beta d_2(x,a) \lt \beta\delta = \delta_1$, so by choice of $\delta_1$ we have $d_1(f(x),L)\lt \frac{\varepsilon}{\alpha}$. Then
$$d_2(f(x),L) \leq \alpha d_1(f(x),L) \lt \alpha\left(\frac{\varepsilon}{\alpha}\right) = \varepsilon.$$
In summary, given $\varepsilon\gt 0$ there exists $\delta\gt 0$ such that if $0\lt d_2(x,a)\lt \delta$, then $d_2(f(x),L)\lt \varepsilon$.
Thus, if the metric you pick is strongly equivalent to the usual metric, then limits are the same and therefore derivatives are the same.
In fact, it is enough that the metrics be topologically equivalent, that is, that for each $x$ and each $\delta\gt 0$ there exist $\delta_1\gt 0$ and $\delta_2\gt 0$ such that 
$$\{y\in\mathbb{R}\mid d_2(x,y)\lt \delta_2\}\subseteq \{y\in\mathbb{R}\mid d_1(x,y)\lt \delta\}$$
and
$$\{y\in\mathbb{R}\mid d_1(x,y)\lt \delta_1\}\subseteq \{y\in\mathbb{R}\mid d_2(x,y)\lt \delta\},$$
will suffice, because the topology determines convergence of sequences, and limits may be defined in terms of sequences; and two metrics are topologically equivalent if and only if they induce the same topology. (Topologically equivalent is weaker than strongly equivalent in general, because you are not requiring uniform "scaling" factors).
