what is the derivative of r = t squared/8 + 1 ?

Hello,

Among my collection of Maths books, I have found a great AS Maths book which is, at last, really clearly explaining **Applications of Differentiation** to me.

But I have just reached a part, in a worked example, that I don’t understand.

The question involves a basic formula: r = t2/8 + 1, (the "t2" bit means "t squared") which gives you the radius of an ink spot, the number of seconds after it first appears.

I don’t understand the last bit of the example. They ask you to find “**the rate of increase of the radius after 4 seconds”.**

They say you need to differentiate, I understand that. But I don’t understand how the derivative that they suggest can be right, as I thought you **ignored numbers**, eg, 2, 3, etc when finding the derivative.(I understand by the way, why numbers are ignored, it’s because eg, 8 can be considered as 8xo etc) But they have **KEPT THE EIGHT in their derivative, and I don’t understand why**.

**Their derivative is:**

**Dr/dt** = **2t/8** = **t/4, **

( based on the original function of: r = t2/8 + 1,_("t squared over 8, plus 1)

And so, they say, after 4 seconds the rate of increase is 4/4 = 1 cm per second.

Is their derivative correct? And if it is, I would be grateful if anyone could explain to me **why the 8 in the original equation is not just ignored when calculating the derivative?**

1/8 is a co-efficient of t squared!

Thanks very much for your great answers e-pi, Prove it, and Halls of Ivy, which I understand perfectly. That is so interesting, the thing about 1/8 being a co-efficient of t squared.It's so helpful to have such useful advice so quickly, it's tiresome when one little thing like that holds you up, thank you very much.