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I am able to complete the action, however, I am not sure what it really means in calculus to differentiate. I'm just curious as to why I have to to do this.

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  • $\begingroup$ Related: math.stackexchange.com/questions/679848/…. $\endgroup$
    – HDE 226868
    Nov 9, 2015 at 3:49
  • $\begingroup$ To differentiate a function means to find its rate of change function. Here's an example: if you've got a function $t\mapsto r(t)$ that describes the position of a car at any time $t$, then the derivative of this function, $t\mapsto r'(t)$, tells you the rate of change -- i.e. speed -- of the car at any time $t$. $\endgroup$
    – user137731
    Nov 9, 2015 at 3:52
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    $\begingroup$ Perhaps one of the answers to these questions will help you: 1, 2, and 3. $\endgroup$
    – user137731
    Nov 9, 2015 at 4:01

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Differentiation is finding the slope. The derivative represents how fast something is changing at an instant - the derivative of position (with respect to time) is speed, for instance. The derivative of speed (with respect to time) is acceleration. Derivatives can tell you how fast you're making a profit based on the amount of money that you have, how fast something is filling up based on its volume, or how fast anything changes given a function representing the value of that anything. Calculus is the mathematics of change.

This is what the definition of the derivative is:

$$\frac{df(x)}{dx}=f'=\color{magenta}{\lim_{h\to 0}}\frac{\color{red}{f(x+h)-f(x)}}{\color{blue}{(x+h)-x}}$$

The slope of a line is $\frac{\color{red}{\Delta y}}{\color{blue}{\Delta x}}$; we want to find the slope as the second point gets closer and closer to the first, which we use a limit for (in pink).

Usually you'll see textbooks simplify the bottom part to just $h$ to get $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$

which is simpler but obscures the reason for the definition.

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  • $\begingroup$ Thank you very much! The color coded formula was a very nice addition, as well. $\endgroup$ Dec 2, 2015 at 19:19
  • $\begingroup$ @Mone: No problem, glad I could help! Hopefully the derivative makes sense to you now c: $\endgroup$
    – Deusovi
    Dec 2, 2015 at 19:36
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To differentiate a function is to calculate its derivative function. This action only makes sense on a function that is differentiable- one of the prerequisites for this is for the function to be continuous. Moreover, the differential quotient $$ \lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$ must exist on all $x\in(a,b)$. In this case we say $f$ is differentiable on $(a,b)$. If it exists for all $x\in\mathbb{R}$ we can simply say that the function is differentiable.

But, enough formality- that's probably not why you asked the question. This information is available in any decent text book. The intuition for differentiation is that the resulting function $f'(x)$ gives you the instantaneous rate of change of $f$ at the point $(x,f(x))$.

Why is this useful? The simplest applications (I know of) are present in the sciences- such as physics. For example, suppose we have a function of displacement for an object as a function of time $D(t)$- the distance from some point as a function of time. The derivative function: $$ D'(t)=V(t)$$ represents the velocity of the object. Moreover: $$ D''(t)=V'(t)=A(t)$$ is the acceleration, or the instantaneous rate of change of the velocity. To my mind, this is the most concrete example of a rate of change. Other examples are prevalent in Chemistry, Economics, the Biological Sciences, and any form of mathematical modeling.

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  • $\begingroup$ You're correct- thanks. $\endgroup$ Nov 9, 2015 at 4:13
  • $\begingroup$ Quick question, is the following wording for describing the derivative of a function as a rate of change correct: the derivative of a function at a point is the rate of change of the function value with respect to its argument at that point? (For example if we have something like $y=f(x)$, then $f'(x_{0})$ is the rate of change of $y$ (the function value) with respect to $x$ (the argument) at the point $x_{0}$ - just so you know where I'm coming from.) $\endgroup$ May 18, 2021 at 20:16
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    $\begingroup$ Yes, that's probably the most precise way to express this statement. I guess people usually just say that it's the rate of change of the function, but really they mean the rate of change of its output with respect to the input. $\endgroup$ May 18, 2021 at 21:11
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To differentiate a function $f$ with respect to the variable $x$ is to find the rate of change, or equivalently to find the slope of the tangent line at any point $P$.

For instance, let $y=f(x)=x^2$, then the derivative is $f'(x)=\frac{\mathrm{d}f}{\mathrm{d}x}=2x$, and this tells us that for any point with $x$-coordinate $x_0$ the slope of the line tangent to the graph of $y=f(x)$ at $x_0$ is simply $2x_0$. So, the slope of the tangent line at $x=1$ is $2(1)=2$, for example.

Another way of looking at this is in terms of speed or velocity. If we have some particle whose motion, which we assume to be in one dimension, is described by some function of time, $x(t)$ (note that here $x$ is the function of the variable $t$, where $x$ is to represent the position and $t$ is to represent time), then we can find the velocity (in the one dimensional case, this is just speed, but with a sign depending on whether the motion is in the direction of positive or negative $x$) by differentiating.

In other words, the speed at time $t$, $s(t)$ is given by

$$s(t)=|v(t)|=|x'(t)|=|\frac{\mathrm{d}x}{\mathrm{t}}|=|\dot{x}|$$

where the notation $\dot{x}$ is just a way of saying the derivative of $x$ with respect to time, $t$.

So, say we have a particle whose motion is linear given by $x(t)=2t+1$, then the velocity is given by the derivative $v(t)=x'(t)=2$. So, the velocity is constant: for any point in the particle's motion, i.e. at any time $t_0$, the velocity is simply $2$, in whatever units are appropriate (say $\mathrm{m/s}$ if the position were given in meters).

There are many other ways at looking at the derivative, though. Formally, we can say

$$f'(x)=\lim_{\Delta x \to 0} \frac{f(x + \Delta x)-f(x)}{\Delta x}$$

We can also look at the derivative as either something that operates on a function, or a function itself. In particular, the derivative maps some function $f(x)$ to $f'(x)$, i.e. $f': f(x) \mapsto f'(x)$, for instance in the case above we have $f: 2x \mapsto 2$. We can extend the derivative function beyond $f: \mathbb{R} \to \mathbb{R}$, though, and you will see that in multivariable calculus and complex analysis.

I cannot imagine this question is unique (that is, I am sure it has been asked on here before), but hopefully this will serve you well.

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Let me address the "I'm just curious as to why I have to to do this" part of your question. That is, why you might want to do it.

The typical practical application is finding maxima/minima of functions. At the maximum/"top" or minimum/"bottom" of a function, it's going up on one side and down on the other. So it's "flat" at the very top/bottom, i.e., its slope/derivative is zero. So you calculate the derivative, set that expression to zero, and solve for the independent variable.

Example: You're a food producer and need to manufacture tin cans that contain a volume $V$ (say, one quart) to distribute your product. So $V=\pi r^2h$ ($r$=radius, $h$=height of can), assuming the usual cylindrical shape. But tin costs money, so you want to choose the $r,h$ that minimizes the area $A=2\pi rh+2\pi r^2$ of the can for the given volume $V$.

Substituting $h=V/{\pi r^2}$ from the $V$-equation, we get $A=2V/r+2\pi r^2$. And now we do our derivative $$\frac{dA}{dr}=-2V/r^2+4\pi r$$ which shows us how $A$ changes as we change $r$ (always keeping $V$ constant). The $r$ at which $A$ doesn't change will be the $r$ for our desired minimum $A$, $$\begin{array}{rcl} 0 &=& -2V/r^2 + 4\pi r\\ r^3 &=& V/2\pi\end{array}$$ So that's 5 minutes work that saves you millions of dollars if you're, say, Campbell Soups.

Extra credit: You have to solder the seams of the can, one seam up the side and two seams around the top and bottom, so that's $h+4\pi r$. Solder costs $c_s\ \mbox{cents/cm}$ and tin costs $c_t\ \mbox{cents/cm}^2$. Okay, now what's the $r$ for the cheapest can containing $V$??? Your turn...

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  • $\begingroup$ In your formula $A = 2V/r + \pi r^2$, the second term should be $2\pi r^2$, making the second term in the formula for $dA/dr$ equal to $4\pi r$, so at the minimal $A$ we have $r^3 = V/(2\pi)$. Since $h = V/(\pi r^2)$ in general, at the $r$ where $A$ is minimized we have $h = Vr/(\pi r^3) = Vr/(V/2) = 2r$, so $h$ is equal to $2r$. That is, the height of the can with minimal surface area having volume $V$ is equal to the diameter of the top of the can. This is not at all what most real soup cans look like, so your comment about this problem saving Campbell Soup millions of dollars is wrong. $\endgroup$
    – KCd
    Nov 9, 2015 at 5:06
  • $\begingroup$ At the same time, it would be nice to how some companies ever really use calculus to optimize the size of the products they manufacture. $\endgroup$
    – KCd
    Nov 9, 2015 at 5:07
  • $\begingroup$ Thanks for pointing out my factor-of-two blunder. I've edited the answer to fix it, so everything's okay now (I hope:). And you're obviously right that soup cans aren't really manufactured in this optinimzed way -- I suppose their geometry is chosen more for marketing purposes. Nevertheless, I think this example still very well serves its illustrative purpose to anwer the op's question about the usefulness of derivatives. But please post a better example. $\endgroup$ Nov 10, 2015 at 7:55
  • $\begingroup$ I agree this is a nice example for using the derivative. So many elementary examples boil down to maximizing or minimizing a quadratic polynomial $ax^2 + bx + c$, and for that you can use ideas related to the quadratic formula to see the max/min is at $x = -b/2a$ with no need for derivatives. $\endgroup$
    – KCd
    Nov 10, 2015 at 7:59

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