# Where am I going wrong?

• Aug 4th 2012, 04:15 PM
astuart
Where am I going wrong?
I'm not sure what I'm doing wrong here...

S(t) = 20,000(1 + e^{-0.5t})

Find the rate of change when t=1

S'(t) = 20,000(1 + e^{-0.5t}) d/dx(-0.5t)
= -10,000(1+e^{-0.5t})

S'(1) = -10,000(1+ e^{-0.5[1]})
= -10,000(1 + (0.606530659)
= -10,000(1.606530659)
= -16,065

Apparently I should be getting -6,065 instead. I realize that this has something to do with the fact that the '1 + ...." is still in the expression, when it shouldn't be, but I'm not sure how I should be getting rid of it..
• Aug 4th 2012, 04:30 PM
emakarov
Re: Where am I going wrong?
According to which rules of differentiation do you think that $\displaystyle (1+e^{f(t)})'=(1+e^{f(t)})f'(t)$?
• Aug 4th 2012, 04:41 PM
astuart
Re: Where am I going wrong?
Quote:

Originally Posted by emakarov
According to which rules of differentiation do you think that $\displaystyle (1+e^{f(t)})'=(1+e^{f(t)})f'(t)$?

I'm going by the chain rule, according to my text-book. From what I understand by my textbook, if f(x) in $\displaystyle e^f(x)$ is differentiable, then the derivative of $\displaystyle e^f(x) = e^f(x) * dx f(x)$.

I realize I'm going wrong here, and it's a bit of a pain because I'm doing this via distance education, which has been lacking at the best of times. I'd be stuffed without this forum.
• Aug 4th 2012, 04:53 PM
emakarov
Re: Where am I going wrong?
You are right about $\displaystyle (e^{f(t)})'$ (up to some typos). What about $\displaystyle \left(1+e^{f(t)}\right)'$? Do you know that $\displaystyle (g(t) + h(t))' = g'(t) + h'(t)$ and $\displaystyle c' = 0$ for a constant c?
• Aug 4th 2012, 05:08 PM
astuart
Re: Where am I going wrong?
Quote:

Originally Posted by emakarov
You are right about $\displaystyle (e^{f(t)})'$ (up to some typos). What about $\displaystyle \left(1+e^{f(t)}\right)'$? Do you know that $\displaystyle (g(t) + h(t))' = g'(t) + h'(t)$ and $\displaystyle c' = 0$ for a constant c?

Hmm, I was aware of it, I Remember that rule from when I was doing derivatives without an exponential. I didn't think of it during this question however.

So, I'm guessing it should be something more like this..?

$\displaystyle S'(t) = 20,000(1 + e^{-0.5t}) * dx (0.5t) = -10,000(e^{-.5t})$

Due to S'(t) removing the constant value from the equation?
• Aug 4th 2012, 05:19 PM
emakarov
Re: Where am I going wrong?
What I can suggest is rewriting the derivative one step at a time. At each step do a single transformation and be sure to identify which of the rules given here you are using.

Quote:

$\displaystyle S'(t) = 20,000(1 + e^{-0.5t}) * dx (0.5t)$
I still don't understand how you obtained the right-hand side. Also, the notation $\displaystyle dx (0.5t)$ is incorrect. Even $\displaystyle dt(0.5t)$ does not mean the derivative of 0.5t.
• Aug 4th 2012, 05:29 PM
astuart
Re: Where am I going wrong?
Quote:

Originally Posted by emakarov
What I can suggest is rewriting the derivative one step at a time. At each step do a single transformation and be sure to identify which of the rules given here you are using.

I still don't understand how you obtained the right-hand side. Also, the notation $\displaystyle dx (0.5t)$ is incorrect. Even $\displaystyle dt(0.5t)$ does not mean the derivative of 0.5t.

I obtained the right side because I thought that S'(t) = S(t) * f '(x).

If f(x) = -0.5t, then f '(x) = -0.5 ?
• Aug 4th 2012, 05:37 PM
emakarov
Re: Where am I going wrong?
Quote:

Originally Posted by astuart
I obtained the right side because I thought that S'(t) = S(t) * f '(x).

First, it should be f(t), not f(x). I assume that f(t) denotes the exponent. Second, S'(t) = S(t) * f '(t) is false because it is not an instance of the chain rule. The correct instance would be R(f(t))' = R'(f(t)) f'(t) where R(x) = 20,000(1 + e^x). Also, it may be easier to factor 20,000 through first.