What is the second derivative for [f(x) = 1/(1+e^-x)]?
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
I solved it ! but I have different solution!!
f''(x) =[ (-e^-x)(1+2e^-x) ] / [1-e^-x]^4
To find the second derivative of f(x), we will first need to find the first derivative, and then take the derivative of that result.
Let's start by finding the first derivative.
Step 1: Find the derivative of the function f(x) with respect to x.
f(x) = 1/(1+e^-x)
To differentiate f(x), we can use the quotient rule:
[f(x) = 1/(1+e^-x)]
f'(x) = [(1)(d/dx(1+e^-x)) - (1+e^-x)(d/dx(1))]/(1+e^-x)^2
Step 2: Find the derivative of (1+e^-x) and (1)
Since the derivative of a constant is 0, the derivative of (1+e^-x) with respect to x is simply the derivative of e^-x, which is -e^-x.
The derivative of (1) with respect to x is 0.
f'(x) = [(1)(-e^-x) - (1+e^-x)(0)]/(1+e^-x)^2
Step 3: Simplify the expression
f'(x) = -e^-x/(1+e^-x)^2
Now that we have the first derivative, we can proceed to find the second derivative.
Step 4: Find the derivative of f'(x)
To differentiate f'(x), we can use the quotient rule again:
f''(x) = [(d/dx(-e^-x))(1+e^-x)^2 - (-e^-x)(d/dx(1+e^-x)^2)]/(1+e^-x)^4
Step 5: Find the derivative of (-e^-x) and (1+e^-x)^2
The derivative of (-e^-x) with respect to x is simply e^-x.
To find the derivative of (1+e^-x)^2, we can use the chain rule. Let u = 1+e^-x, then apply the power rule:
d/dx(u^2) = 2u * (d/dx(u))
d/dx(1+e^-x)^2 = 2(1+e^-x) * (d/dx(1+e^-x))
Step 6: Simplify the expression
f''(x) = [e^-x(1+e^-x)^2 - (-e^-x)2(1+e^-x)]/(1+e^-x)^4
f''(x) = [e^-x(1+e^-x)^2 + 2e^-x(1+e^-x)]/(1+e^-x)^4
Simplifying the expression further, we get:
f''(x) = [e^-x(1+e^-x) + 2e^-x]/(1+e^-x)^3
Therefore, the second derivative of f(x) = 1/(1+e^-x) is given by:
f''(x) = [e^-x(1+e^-x) + 2e^-x]/(1+e^-x)^3