Two different functions, the same derivative. Very interesting.

[jasonlr82794]

Hey guys, I had a question about two functions and many more if you look at it right, that have the same derivative. I took the derivative of x^2 and found its derivative is of course 2x. So I decided to take the derivative and x^2+1 and found the same derivative applies? How can This be? Logically, you would think since the function x^2+1 is building up distance faster than that of x^2 the derivative would at least be slightly greater but it doesn't hold true. And is there anyway that I could look at the function x^2 and logically deduce that its derivative is 2x. Maybe I am not grasping the full concept of the instantaneous derivative. Please help. Thank you.

 

[D H]

Those two functions differ by a constant. Any two functions f(x) and g(x) that differ only by a constant, i.e., f(x)-g(x)=c, will have the same derivative.

 

[jasonlr82794]

Ok, I understand the constant rule and how this applies but if one function is building up distance faster than another wouldn't that imply a slightly faster rate of change?

 

[pwsnafu]

Logically, you would think since the function x^2+1 is building up distance faster than that of x^2 the derivative would at least be slightly greater but it doesn't hold true.

Err, no. The derivative is rate of change, and the rate of change of constants are zero. Why should the gradient of the tangent depend on its position?

And is there anyway that I could look at the function x^2 and logically deduce that its derivative is 2x.

You have the definition of the derivative. That's how everything is "logically deduced": you go back to the definition.

 

[pwsnafu]

Ok, I understand the constant rule and how this applies but if one function is building up distance faster than another wouldn't that imply a slightly faster rate of change?

In your example...they are not. That's why they have the same derivative.

 

[Fredrik]

If you need some physical intuition, compare $x(t)=t^2$ to $x(t)=t^2+1$. That 1 in the second formula is just the starting position (if t=0 is the "start"). It has nothing to do with speed. Note also that in a race between two vehicles that move as described by these functions, the distance between them at time t would be $(t^2+1)-t^2=1$, which is a constant.

 

[jasonlr82794]

Ok, so this one is to Fredrik. I looked at the graph and the starting position like you said was 0,1 and for the other function 0,0. So there distances are building up the same just different positions? I looked on the graphs of these two functions and also noticed it is the same exact graph besided the starting position. I'm pretty sure I got that now but how about if I only knew my derivative and time and had my velocity graph. I integrated this and it only came to the distance answers on the function x^2? Can you explain this?

 

[Office_Shredder]

Ok, so this one is to Fredrik. I looked at the graph and the starting position like you said was 0,1 and for the other function 0,0. So there distances are building up the same just different positions? I looked on the graphs of these two functions and also noticed it is the same exact graph besided the starting position. I'm pretty sure I got that now but how about if I only knew my derivative and time and had my velocity graph. I integrated this and it only came to the distance answers on the function x^2? Can you explain this?

When you integrate a function you have to remember that you can add any constant to that function and get another function which works. This is often denoted with a +C at the end, for example

\int 2x dx = x^2 + C
To remind you that x²+3324.245245 is an equally valid choice of function

 

[jasonlr82794]

Ohhhhhhhhhhhhhhhh, that makes perfect sense. Since they built up distance the same just at different starting positions they will still obtain the same distance no matter what the graph distance says because their derivative is the same. Thank you all for helping me and it really does make a lot more sense now.