Applicability of derivatives and integrals Why do derivatives and integrals work? I understand the concept on how to apply them, but what makes it possible for them to be used? why does taking the derivative of volume equal an object's surface area? I want to know more about why the theorems and rules work the way they do, and how can I apply them to my thinking instead of just knowing that it works.
 A: Integration is simply the sum of small stuffs to form bigger stuffs.


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*You pile up points in a certain direction, you get a line

*You pile up lines in a certain direction , you get a surface

*You pile up surfaces in a certain direction, you get a volume.

*You pile up the cells to form tissues, tissues to form organs, organs to form a complete system.


Derivative is just the opposite of integration and vice-versa. You derive when you want to break something down into smaller pieces.
This explanation may be in some way naïve but that's the way I've understood calculus all my life.
In addition, I think the way they've both worked successfully in many areas of science is from the careful definition of derivation. Given a function $y = f(x)$

The derivative is the rate at which $y$ changes with respect to the change in $x$. If one notes $\text {The change in }y = \Delta y$ and $\text {The change in }x = \Delta x$, the rate (the derivative of $y$ with respect to $x$) is simple $\cfrac {\Delta y}{\Delta x}$.

To get this rate, it is sufficient to $\text {disturb}$ the variable $x$ by an extremely small value, almost $0$ but yet greater than $0$, the value is $\text {The change in }x = \Delta x$. Then after one computes $f(x+\Delta x)$ but note that even if $\Delta x$ is very small, it is not exactly equal to $0$ i.e. $f(x+\Delta x) \ne y$ but $f(x+\Delta x) = y + \Delta y$ where $\Delta y$ is the disturbance on $y$ or change in variable of $y$.
By computing $ y + \Delta y=f(x+\Delta x) $ and some approximation on large powers of $\Delta x$ and $\Delta y$, one can compute the rate $\cfrac {\Delta y}{\Delta x}$.
With this carefully approximated rate, many areas of Science have grown.
A: "why does taking the derivative of volume equal an object's surface area?"
This is best answered in reverse. One calculates the volume of a solid by "summing up"/integrating "thin" slices of volume $A\Delta x$, where $\Delta x$ is the width of a slice and $A$ is the cross-sectional area: $V =\int A dx$. Think of using circular cross-sections to get the volume of a sphere. 
I'm going to be intentionally sloppy here since being mathematically precise can sometimes obscure conceptual understanding: "too much mathematical rigor causes rigor mortis". 
Recall that differentiation and integration are inverses of one another, right? Well, if you differentiate the formula $V =\int A dx$ you get the answer to your question: 
$$\frac{dV}{dx}=\frac{d}{dx} \int A dx =A$$
