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 March 2nd, 2019, 04:17 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 541 Thanks: 82 Implicit function $\displaystyle s=s(t) \; \;$ , $\displaystyle s=e^{-st}$ . Find $\displaystyle s’(t)$ .
March 2nd, 2019, 07:55 AM   #2
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 Originally Posted by idontknow $\displaystyle s=s(t) \; \;$ , $\displaystyle s=e^{-st}$ . Find $\displaystyle s’(t)$ .
Use the chain rule.

I'll get you started:
$\displaystyle \dfrac{ds}{dt} = e^{-st} \cdot \dfrac{d}{dt} ( -st)$

-Dan

 March 2nd, 2019, 01:17 PM #3 Global Moderator   Joined: May 2007 Posts: 6,770 Thanks: 700 s=e^{-st} is strange. You should replace one s by another symbol. Thanks from idontknow Last edited by mathman; March 2nd, 2019 at 01:22 PM.
 March 2nd, 2019, 01:32 PM #4 Senior Member   Joined: Dec 2015 From: somewhere Posts: 541 Thanks: 82 Saw it in a old book (calculus 3) . This is the reason I posted it , by the graphing calculator the function is continuous and maybe it is possible that s’ exists . And what I got is $\displaystyle s’=\frac{s^2 }{s\ln s -1}$ .
March 2nd, 2019, 01:59 PM   #5
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Quote:
 Originally Posted by idontknow Saw it in a old book (calculus 3) . This is the reason I posted it , by the graphing calculator the function is continuous and maybe it is possible that s’ exists . And what I got is $\displaystyle s’=\frac{s^2 }{s\ln s -1}$ .
How did you get that?

Picking up where I was before:
$\displaystyle \dfrac{ds}{dt} = e^{-st} \cdot \dfrac{d}{dt}(-st)$

$\displaystyle \dfrac{ds}{dt} = e^{-st} \cdot \left ( -\dfrac{ds}{dt} t - s \right )$

$\displaystyle \dfrac{ds}{dt} = - t e^{-st} \cdot \dfrac{ds}{dt} - s e^{-st}$

Now solve for $\displaystyle \dfrac{ds}{dt}$. There will be no logarithms.

-Dan

March 2nd, 2019, 03:57 PM   #6
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Quote:
 Originally Posted by idontknow Saw it in a old book (calculus 3) . This is the reason I posted it , by the graphing calculator the function is continuous and maybe it is possible that s’ exists. And what I got is $\displaystyle s’=\frac{s^2 }{s\ln s -1}$ .
Wait, are you trying to eliminate t from your answer? Then you are close. It's
$\displaystyle \dfrac{ds}{dt} = \dfrac{s^2}{\ln(s) - 1}$.

Normally you wouldn't do this unless explicitly told to.

-Dan

Last edited by skipjack; March 3rd, 2019 at 04:11 PM.

March 2nd, 2019, 05:37 PM   #7
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Quote:
 Originally Posted by mathman s=e^{-st} is strange.
It's a functional equation.

March 3rd, 2019, 02:26 PM   #8
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Quote:
 Originally Posted by greg1313 It's a functional equation.
It is also a non-explicit function.

Last edited by skipjack; March 3rd, 2019 at 04:25 PM.

 March 3rd, 2019, 04:24 PM #9 Global Moderator   Joined: Dec 2006 Posts: 20,759 Thanks: 2138 $s = e^{-st} \implies t = -\ln(s)/s$, so $dt/ds = (\ln(s) - 1)/s^2$. Hence $\displaystyle s' = \frac{s^2}{\ln(s) - 1}$. For obvious reasons, the cases $s < e$ and $s > e$ need to be considered separately. Thanks from topsquark and idontknow

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