# Intuition behind the derivative of are of a square? How to properly use the derivative ?

If I derive the formula $$S=16t^2$$, where S denotes the distance and t denotes time I get $$ds/dt= 32t$$. This in return give me a formula for the speed of the object at any time t. However if we were to do the same thing to the formula $$A=S^2$$, where A denotes the Area of the square and S denotes the length what would the derivative tell me. I know that its the instantaneous rate of change of the Area with respect to a small change in the length S but is there another meaning behind it? In addition I would like to understand how do I use the derivative. For example if we were to derive the equation A=S^2 we would get da/ds= 2S. What this is telling me is that if I increase the length by ds the change in Area da is 2S. In consequence if I take a square of size 6 and decide to increase its length by 1 the Area should increase by 2*6. This however doesnt not happen. If A1= 36, A2= 49. The difference in the increase Area is 13 not 12. So am I missing something here ?

Say that the side area changes by $\Delta s$. The change in area of the square is $$(s+\Delta s)^2-s^2=2s\Delta s + (\Delta s)^2.$$ If $\Delta s$ is small enough we can approximate the change in area by $2s\Delta s$. In your case the change in area is $$2\times6\times 1+1^2=13$$
Yes, it happens, but you shouldn't take $1$ but rather $0.01$. Then really $6.01^2$ is close to $36+2\times 6\times 0.01$. This picture might help.
The idea is that if $dS$ is small, then the "neglected" part is even much smaller.
• But that's the point of the theory of derivatives! It is only infinitesimal, that is, these formulas are only valid in the limit $ds\to 0$. Look up the definitions in your book. It's the same as with velocity: to compute the velocity in one time instance you need to measure the distance one milisecond before / after and divide by the tiny fraction of time. – Peter Franek Jun 23 '16 at 17:37