Suppose that we have a square sheet of edge length $L$. Its area $A=L^2$.

Differentiating $A$ w.r.t. L, we get

I do understand what it means to differentiate, graphically, it gives you the slope of the tangent at a point on the graph. But now, when I think of what differentiating means in the context of Area and length, it doesn't make any sense to me at all. What does $2L$ signify?

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    $\begingroup$ Rate of change of area with respect to length. $\endgroup$ – evil999man Apr 30 '14 at 11:53

Try to draw a square $ABCD$ with side equal to $L$. Now draw a slightly bigger square $AB'C'D'$ with side length $L+\Delta L$ (such that $DD'=BB'=\Delta L$). Now look at the $\Gamma$-like shape cut from $AB'C'D'$ by $ABCD$, you can split it into three parts: two thin rectangles $L\times \Delta L $ and one small square $\Delta L\times \Delta L $.

Now the derivative is in quite simpified terms "the difference of value of the function over the change of argument", so basically when you increase the side length by $\Delta L$, then the surface increases by $2L\Delta L$ and a negligeble term $(\Delta L)^2 $.

One can also say that $2L$ signifies the permeter of the part of the square that got inflated.

  • $\begingroup$ Did you mean Perimeter? $\endgroup$ – Shaurya Gupta Apr 30 '14 at 11:55
  • $\begingroup$ I understood most of it...But to understand it completely, I think I need to know what it means for $2x$ to be the derivative of $x^2$. $\endgroup$ – Shaurya Gupta Apr 30 '14 at 12:00
  • $\begingroup$ Is $2L$ the change in Area for unit change in length? $\endgroup$ – Shaurya Gupta Apr 30 '14 at 12:01
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    $\begingroup$ No, but for an infinitely small $\Delta L$ the change in area is $2L\Delta L$. The $2L$ signifies how much the area would change relative to an infinitely small change in length. Imagine the graph for $x^2$ - for a certain $x$, if you'd go an infinitely small bit to the left or right, the slope would be $2x$, that's what it means. If the length becomes a very little bit bigger or smaller, it also changes the area by $2L *$ a very little bit. $\endgroup$ – EagleV_Attnam Apr 30 '14 at 13:59
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    $\begingroup$ So you could write $\Delta A = 2 L \Delta L + (\Delta L)^2$. Dividing by $\Delta L$ gives $\frac{\Delta A}{\Delta L} = 2L + \Delta L$. If you now make $\Delta L$ very, very small (mathematically, take the limit $\Delta L \to 0$), you will get $2L$ on the right hand side. The left hand side is the definition of $\frac{dA}{dL}$. $\endgroup$ – CompuChip Apr 30 '14 at 14:07

Consider this picture:

Derivative of area of is square is twice the length of its side

Here, the green square is the square of area $A=L^2$ and red line is its increase.

When you increase the length $L$ by $dL$, the area $A$ gets increased by $2LdL$. So, to answer your question, significance of $2L$ is that it is the length of the red line on the picture ($dL$ is its width).

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    $\begingroup$ This may be confusing since the area of the red part is actually 2LdL + dL^2... $\endgroup$ – Heisenberg Apr 30 '14 at 16:33
  • $\begingroup$ @Anh Indeed, but $(dL)^2$ part becomes infinitely smaller than $2LdL$ in the limit when $dL$ becomes infinitesimal, so it can be safely ignored. It matters for the second derivative though. $\endgroup$ – Danijel Apr 30 '14 at 17:01
  • $\begingroup$ @Danijel, you should add that to your answer, to make it more complete. Still, +1 for the picture $\endgroup$ – JMCF125 May 1 '14 at 12:38

Thinking of the derivative graphically as the slope of the tangent is just one way to understand the meaning of the derivative. It's the most common, because it's how the derivative is motivated in most introductory calculus courses. But the meaning and value of the idea of a derivative is much deeper. The derivative measures the rate at which something changes. That's worth thinking about before you start with graphs and formulas. Here are some examples.

Suppose you're driving. Then the distance you've traveled changes as time goes by. If you're driving along at a constant 30 miles per hour then the distance increases by 30 miles for each hour of travel. The derivative of the distance is the rate: 30 miles per hour.

That's an easy example because the rate of travel is constant. Calculus was invented to handle situations where the rate is itself changing. For example, if you start from a red light and accelerate up to the legal speed limit of 30 miles per hour then your speed is changing. The derivative of the speed is the rate at which you're speeding up - the acceleration. You might measure that in (miles per hour) per second.

In economics, the number of customers for your product depends on the price you charge. When you raise the price, fewer people will buy from you. The derivative of the number of customers is the rate at which you lose them, measured in (customers lost) per (dollar increase in price). In this case the derivative is negative.

Populations change over time. For microorganisms you might choose to measure time in hours. Then the derivative of the population is the number of new organisms per hour. Then things get interesting, because the number of new organisms per hour depends on the population - the more organisms you have, the more of them there are to reproduce. So the derivative of the population, measured in new organisms per hour, is the product of the number of organisms and the birth rate. That means the derivative of the population (as time goes on) is proportional to the population. That leads to exponential growth.

You can describe the derivative of a graph of the function y = f(x) the same way. Here the height y changes as the value of x changes. The steeper the graph (at any particular point) the larger the change in y for any particular small change in x. The rate at which y changes is the derivative. You have to think only about small changes in x since the graph is a curve, whose steepness varies from place to place. As long as the change in x is small, the curve nearly matches the tangent, whose slope is just the rate of change you care about. (It's taken mathematicians centuries of work to make precise sense of the idea expressed roughly as "if you change x by just an infinitesimal amount then the curve and the tangent are the same".)

Now think about the question you asked. The area of a square depends on the length of its side. The derivative measures the rate at which the area changes when the side changes, measured in units like (square centimeters of area) per (centimeter of side). @TZakrevskiy 's answer above explains why that's just twice the side length. Here's an analogous question: explain why when you grow a circle of radius r the area changes at the rate 2 pi r.

I wish there were more time and more incentive to spend time in calculus classes on these ideas, rather than rushing to the rules and formulas for derivatives (and integrals).

  • $\begingroup$ So derivative gives us the change in something "per what"? $\endgroup$ – Shaurya Gupta Apr 30 '14 at 15:33
  • $\begingroup$ @shauryagupta Yes. The "what" in "per what" depends on how you frame your question - it's the independent variable. You could think about how the amount of gas you use depends on how far you travel, or about how it depends on the speed at which you travel (auto engine efficiency falls off at really high speeds). You'd have different derivatives in these examples. $\endgroup$ – Ethan Bolker Apr 30 '14 at 15:59
  • $\begingroup$ So if I differentiated amount of gas w.r.t. distance, I would get the change per how much of distance? $\endgroup$ – Shaurya Gupta Apr 30 '14 at 16:05
  • $\begingroup$ @shauryagupta Yes. You would get the rate of change of gas used, in units (gallons of gas) per (miles traveled). In the other problem you would get the rate in (gallons of gas used) per (speed at which you travel). At high speeds an increase in the speed would cause an increase in the gas used (positive derivative). At low speeds an increase in the speed would cause a decrease in gas used (negative derivative). The derivative would be zero at the optimal speed. $\endgroup$ – Ethan Bolker Apr 30 '14 at 16:16
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    $\begingroup$ @shauryagupta Indeed it is. That's what makes calculus hard, and why it took so long for mathematicians to invent it. $\endgroup$ – Ethan Bolker Apr 30 '14 at 16:19

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