March 2nd, 2019, 04:17 AM  #1 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68  Implicit function
$\displaystyle s=s(t) \; \;$ , $\displaystyle s=e^{st}$ . Find $\displaystyle s’(t)$ . 
March 2nd, 2019, 07:55 AM  #2 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,079 Thanks: 845 Math Focus: Wibbly wobbly timeywimey stuff.  
March 2nd, 2019, 01:17 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,704 Thanks: 670 
s=e^{st} is strange. You should replace one s by another symbol.
Last edited by mathman; March 2nd, 2019 at 01:22 PM. 
March 2nd, 2019, 01:32 PM  #4 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 
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  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,079 Thanks: 845 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
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  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,079 Thanks: 845 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle \dfrac{ds}{dt} = \dfrac{s^2}{\ln(s)  1}$. 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 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,923 Thanks: 1123 Math Focus: Elementary mathematics and beyond  
March 3rd, 2019, 02:26 PM  #8 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68  It is also a nonexplicit 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: 20,373 Thanks: 2010 
$s = e^{st} \implies t = \ln(s)/s$, 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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