# How do you get the derivative of this?

1. Sep 6, 2006

### GrifteR150

This is for my Circuits class, I am supposed to find the current i.

If q = 5te^(-10^(3) t) Coulombs

Find i (current) the formula for i is: i = dq/dt

So what I did was try to find the derivative of q, which is dq

So I brought out the constant 5, and I am left with te^(-10^(3) t)

and I recognize this as a product of two functions so I tried the product rule.

I let my first function f = t, and my second function g = e^(-10^(3) t)

so f ' = 1 and g ' (using the chain rule) i got (e^(-10^(3) t) x '10^3)

so i plugged it into the product role (fg)' = fg' + gf' and i get a weird answer.

The given answer in the back of the book is :

i = 5(1-10^3 t)e^(-10^3 t) Amps

2. Sep 6, 2006

### TD

Perhaps you just need to simplify? What does your current answer look like?

PS: we have a homework section

3. Sep 6, 2006

### chroot

Staff Emeritus
Your general approach is correct. g' = e^(-10^(3) t) * -10^3

When you plug this into the product rule, you get a "weird answer?" Are you sure that your weird answer is not, in fact, the correct answer in unsimplified form?

- Warren

4. Sep 6, 2006

### George Jones

Staff Emeritus
It looks to me that if you make a small correction in your calculation of g', then you will get the same answer as the book.

5. Sep 6, 2006

### HallsofIvy

I presume that is a typo and it should be (e^(-10^(3)t) x (-10^3)) not
'10^3.

How weird? You just said $f= t, g'= -10^3 e^{-10^3t}, g= e^{-10^3t}, f'= 1$ so $fg'+ f'g= -10^3 te^{-10^3t}+ e^{-10^t}$
What do you get if you factor $e^{-10^3 t}$ out of that?

Last edited by a moderator: Sep 6, 2006