# The derivative

• Dec 3rd 2007, 09:04 AM
Coach
The derivative
I've tried it, but I always get the wrong answer, I get \$\displaystyle h'(t)=s^2+s \$

Thank you!
• Dec 3rd 2007, 09:06 AM
Jhevon
Quote:

Originally Posted by Coach
I've tried it, but I always get the wrong answer, I get \$\displaystyle h'(t)=s^2+s \$

Thank you!

s is a constant, is it not? what is the derivative of a constant?
• Dec 3rd 2007, 09:09 AM
Coach
0, I guess.

but according to my book the answer is \$\displaystyle h'(t)=s^2 \$
• Dec 3rd 2007, 09:10 AM
Jhevon
Quote:

Originally Posted by Coach
0, I guess.

but according to my book the answer is \$\displaystyle h'(t)=s^2 \$

because the derivative of the lone s is zero, as you just said...why are you guessing by the way?
• Dec 3rd 2007, 09:54 AM
Coach
Ok, but I thought, that since h is a function of t, I should only derivate t, and since t becomes 1, then s remains.

I am guessing, because I usually get it wrong, and I am not sure about it(though I am sure that the derivative of a constant is zero).

But thank you so much!
• Dec 3rd 2007, 06:15 PM
Jhevon
Quote:

Originally Posted by Coach
Ok, but I thought, that since h is a function of t, I should only derivate t, and since t becomes 1, then s remains.

not when s is by itself. the derivative of h(t) = 2t + 1 = 2 for instance. the t goes to one, you leave it's coefficient, but the 1 goes to zero, because it is a constant