# Math Help - Where am I going wrong?

1. ## 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..

2. ## Re: Where am I going wrong?

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

3. ## Re: Where am I going wrong?

Originally Posted by emakarov
According to which rules of differentiation do you think that $(1+e^{f(t)})'=(1+e^{f(t)})f'(t)$?
I'm going by the chain rule, according to my text-book. From what I understand by my textbook, if f(x) in $e^f(x)$ is differentiable, then the derivative of $e^f(x) = e^f(x) * dx f(x)$.

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.

4. ## Re: Where am I going wrong?

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

5. ## Re: Where am I going wrong?

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

$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?

6. ## 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.

$S'(t) = 20,000(1 + e^{-0.5t}) * dx (0.5t)$
I still don't understand how you obtained the right-hand side. Also, the notation $dx (0.5t)$ is incorrect. Even $dt(0.5t)$ does not mean the derivative of 0.5t.

7. ## Re: Where am I going wrong?

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 $dx (0.5t)$ is incorrect. Even $dt(0.5t)$ 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 ?

8. ## Re: Where am I going wrong?

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.