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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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 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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 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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 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 Tags function, implicit Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post shaharhada Calculus 2 October 4th, 2018 11:34 PM shaharhada Algebra 2 March 15th, 2018 11:06 AM shaharhada Math 6 December 9th, 2017 03:56 AM Azkoul Calculus 1 August 31st, 2015 07:22 AM MATHEMATICIAN Abstract Algebra 5 February 12th, 2015 09:37 AM

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