Asked by 008
What is the second derivative for [f(x) = 1/(1+e^-x)]?
Answers
Answered by
Reiny
I have just answered a question on this same function when you posted as "009".
Keep the same name please.
In this case I will simplify the function a bit
f(x) = 1/(1 + e^-x) = (1+e^-x)^-1 * e^x/e^x
= e^x/(e^x + 1)
f ' (x) = ( (e^x + 1)(e^x) - (e^x)(ex) )/(e^x + 1)^2
= e^x/(e^x + 1)^2
f '' (x) = ( (e^x + 1)^2 (e^x) - (e^x)(2)(e^x + 1)(e^x) )/(e^x + 1)^3
I will leave it up to you to simplify the top
Keep the same name please.
In this case I will simplify the function a bit
f(x) = 1/(1 + e^-x) = (1+e^-x)^-1 * e^x/e^x
= e^x/(e^x + 1)
f ' (x) = ( (e^x + 1)(e^x) - (e^x)(ex) )/(e^x + 1)^2
= e^x/(e^x + 1)^2
f '' (x) = ( (e^x + 1)^2 (e^x) - (e^x)(2)(e^x + 1)(e^x) )/(e^x + 1)^3
I will leave it up to you to simplify the top
Answered by
008
I solved it ! but I have different solution!!
f''(x) =[ (-e^-x)(1+2e^-x) ] / [1-e^-x]^4
f''(x) =[ (-e^-x)(1+2e^-x) ] / [1-e^-x]^4
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