Can any one explain me what is the use of derivatives in real life. When and where we use derivative, i know it can be used to find rate of change but why?. My logic was in real life most of the things we do are not linear functions and derivatives helps to make a real life functions into linear.

eg converting parabola into a linear function $x^{2}\rightarrow 2x$

but then i find this; derivation of $\sin{x}\rightarrow \cos{x}$

why we cant use $\sin{x}$ itself to solve the equation. whats the purpose of using its derivative $\cos{x}$. Please forgive me if i have asked a stupid questions, i want to improve my fundamentals in calculus.

  • $\begingroup$ I think this should be CW. $\endgroup$ – Joe Jun 20 '12 at 16:49

My example is from the real life situation of war. From experiments in physics we know that the acceleration due to gravity of a particle near the Earth's surface is about $-9.8 \frac{m}{s^2}$.

If you're an artilleryman in an army, you want your artillery shells to hit the enemy, or else theirs may hit you and kill you (game over). So you need to know how to angle your artillerygun and which direction to point it in so that when the shell lands, it blows up your enemy (rather than missing).

All you know is that you can control the direction your cannon points and the angle you fire it in in the air -- after it's fired gravity takes over and that $-9.8 \frac{m}{s^2}$ takes over the situation.

Calculus shows us that if $x(t)$ is the function representing the position of the artillery shell at time $t$, then its first derivative is the shell's velocity and its second derivative is the shell's acceleration. We know acceleration from physics (it's that $-9.8 \frac{m}{s^2}$ we had earlier)! We write this as $x''(t) = -9.8 \frac{m}{s^2}$.

From this equation (involving derivatives) you can calculate (using "integration") the position function $x(t)$ of the particle given the direction you fire it in and the angle you fire it at.

Why do you care about that? Because knowing $x(t)$ will tell you where your shell lands and thus whether your shot will kill the enemy or not. So you can do a quick calculation to determine which direction and which angle to fire in to ensure that your shell hits your target. The side that does this computation first and gets the shell in the air first will kill the other side, helping win the war.

  • $\begingroup$ Minor note: there would probably be air resistance against the shells turning it into a simple DE. $\endgroup$ – Joe Jun 20 '12 at 16:53
  • 2
    $\begingroup$ Of course in real life the gravity is not constant as you go higher in the air either. I think it would be bad pedagogy to include such details when trying to explain the first time to someone who doesn't yet understand calculus. $\endgroup$ – tomcuchta Jun 21 '12 at 0:07
  • $\begingroup$ Yup, I agree completely. :) $\endgroup$ – Joe Jun 21 '12 at 1:34

Yes, the derivative can be used to determine the "rate of change" but more generally can be viewed as a tool to approximate nonlinear functions locally with linear functions. This is true in the case of a real-valued function of a real variable and is the case in higher dimensions such as a surface defined by a multivariable function. In the former case, the linear function is realized as the line tangent at a given point and in the latter as the tangent plane. The derivative is, in essence, the best linear model available for a function in a neighborhood of a point.

The derivative, by providing a mechanism of "local linearization", can turn a hard/intractable problem into a problem of linear algebra which is usually easier to deal with. In real life, the utility comes into play when complex behavior of physical systems is modeled by nonlinear functions which can in then in turn be locally approximated by the derivative.


From another point of view, the derivative represents how one quantity changes as another quantity varies. In many cases, we can construct models by relating quantities to their rates of change, then using tools from differential equations to make predictions. The predictions are usually very accurate, which is why we teach calculus to our high school and college students.


A simple object's position is governed by an equation relating the second derivative of its position to the forces on the object: $F_{net} = mx''(t)$.

Short-run employment in the economy depends linearly on the time derivative of the difference between the expected price level and the actual price level.

The rate at which heat spreads through an object from a constant-temperature source is related to the derivative of temperature with respect to distance.

In simple biological systems, the population is proportional to its time derivative. In slightly more complicated biological systems, the time derivative of the population is proportional to the population multiplied by the difference between the total sustainable population and the population.

This list could be much, much longer, but I think it illustrates the point. Derivatives are useful.


Derivatives are very useful. Because they represent slope, they can be used to find maxima and minima of functions (i.e. when the derivative, or slope, is zero). This is useful in optimization.

Derivatives can be used to estimate functions, to create infinite series. They can be used to describe how much a function is changing - if a function is increasing or decreasing, and by how much. They also have loads of uses in physics. Derivatives are used in L'Hôpital's rule to evaluate limits. Derivatives can even help you graph a function!


There can be also economic interpretations of derivatives. For example, let's assume that there is a function which measures the utility from consumption.

$U(C)$ where $C$ is the consumption. It is straightforward to say that your utility increases with consumption. This means that when you increase your consumption one unit marginally, you will have an increase of your utility, which you can find it by taking derivative of this function. (it is not a partial derivative because the only argument in your function is consumption.) So you will have ;

$\frac{dU(C)}{dC} > 0 $

By intuition, the increase of the utility will be less more you consume. Let's take the simplest example, you have one coke and you drunk it. The second coke you will drink just after will provide you utility but less. The increase of your utility will be less fast more you consume. This one can be formulated by the second derivative of the utility function as follows ;

$\frac{d^{2}U(C)}{dC^{2}} < 0$


protected by Asaf Karagila Dec 11 '14 at 10:27

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