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March 2nd, 2019, 04:17 AM   #1
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Implicit function

$\displaystyle s=s(t) \; \;$ , $\displaystyle s=e^{-st}$ .
Find $\displaystyle s’(t)$ .
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March 2nd, 2019, 07:55 AM   #2
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$\displaystyle s=s(t) \; \;$ , $\displaystyle s=e^{-st}$ .
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
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March 2nd, 2019, 01:17 PM   #3
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s=e^{-st} is strange. You should replace one s by another symbol.
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Last edited by mathman; March 2nd, 2019 at 01:22 PM.
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March 2nd, 2019, 01:32 PM   #4
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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 $\displaystyle s’=\frac{s^2 }{s\ln s -1}$ .
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March 2nd, 2019, 01:59 PM   #5
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Originally Posted by idontknow View Post
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.

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March 2nd, 2019, 03:57 PM   #6
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Quote:
Originally Posted by idontknow View Post
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
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Last edited by skipjack; March 3rd, 2019 at 04:11 PM.
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March 2nd, 2019, 05:37 PM   #7
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s=e^{-st} is strange.
It's a functional equation.
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March 3rd, 2019, 02:26 PM   #8
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It's a functional equation.
It is also a non-explicit function.

Last edited by skipjack; March 3rd, 2019 at 04:25 PM.
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March 3rd, 2019, 04:24 PM   #9
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$s = e^{-st} \implies t = -\ln(s)/s$, 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.
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