derivative^(s/t)
6 answers
derivative e^(s/t)
I don't understand how the answer is(-s/t^2)e^(s/t) and not ((t-s)/t^2)e^(s/t)
Please show
Please show
I have no idea what "derivative^(s/t)" even means.
However, looking at your answer, I suppose that if
y = e^(s/t)
then using the chain rule and the quotient rule,
dy/dt = e^(s/t) * d/dt (s/t)
= e^(s/t) (-s/t^2)
s is just a constant, like 2. If s is also a function of t, then you have
∂y/∂t instead of dy/dt
However, looking at your answer, I suppose that if
y = e^(s/t)
then using the chain rule and the quotient rule,
dy/dt = e^(s/t) * d/dt (s/t)
= e^(s/t) (-s/t^2)
s is just a constant, like 2. If s is also a function of t, then you have
∂y/∂t instead of dy/dt
How did you get (-s/t^2) instead of ((t-s)/t^2 ? I got that answer from using the chain rule and the quotient rule too. (s/t) -->(t*1-s*1)/(t)^2 --> =(t-s)/t^2)
consider
y = e^(2/t)
what is dy/dx in that case?
Is s a constant, or a function of t?
If s is a function, then if
y = e^(s/t)
y' = e^(s/t) (s't - s)/t^2
y = e^(2/t)
what is dy/dx in that case?
Is s a constant, or a function of t?
If s is a function, then if
y = e^(s/t)
y' = e^(s/t) (s't - s)/t^2
I think I get what you mean now. If s is a function, then it is like this dy/dt (s/t)=(t-s)/t^2 right ? But if s is a constant, then it is like this d/dt(s/t)= -s/t^2 right? My teacher didn't teach us this so I'm just want to be sure.
5 answers