I don't get how you get this part:
f' = (1-x)/(1+x)^3
f" = [(-1)(x+1)^3 - 3(1-x)(x+1)^2]/(x+1)^6
= [-(x+1)+3(x-1)]/(x+1)^4 ( I don't understand how you got this!!)
= (-x-1+3x-3)/(x+1)^4
= (2x-4)/(x+1)^4
f" = [(-1)(x+1)^3 - 3(1-x)(x+1)^2]/(x+1)^6
= [-(x+1)+3(x-1)]/(x+1)^4 ( I don't understand how you got this!!)
Looks like division of top and bottom by (x+1)^2 and also -(1-x) = + (x-1)
Thank you, Damon!
You are welcome.
To understand how to derive f" from f', let's go through the steps one by one:
1. You start with the expression for f':
f' = (1-x)/(1+x)^3
2. To find the second derivative, we need to differentiate f'. We'll apply the quotient rule:
The quotient rule states that if you have a function in the form of f(x) = u(x)/v(x), the derivative of f(x) with respect to x is given by:
(f'(x)) = (u'(x)v(x) - v'(x)u(x))/(v(x))^2
If we apply the quotient rule to f', we have:
f"(x) = [(1+x)^3 * (d/dx(1-x)) - (1-x) * (d/dx((1+x)^3))] / ((1+x)^3)^2
3. Now, let's calculate the first derivative of (1-x) and (1+x)^3 to continue simplifying:
d/dx(1-x) = -1
d/dx((1+x)^3) = 3(1+x)^2
4. Plugging these values into the f" equation, we have:
f"(x) = [(1+x)^3 * (-1) - (1-x) * (3(1+x)^2)] / ((1+x)^3)^2
5. Simplifying further:
f"(x) = [(-1)(1+x)^3 - (1-x)(3(1+x)^2)] / (1+x)^6
6. Distributing negative sign:
f"(x) = [-1 * (1+x)^3 - (1-x) * 3(1+x)^2] / (1+x)^6
7. Expanding terms:
f"(x) = [-(1+x)^3 - 3(1-x)(1+x)^2] / (1+x)^6
8. Simplifying further:
f"(x) = [-(1+x) + 3(1-x)] / (1+x)^4
9. Expanding the terms inside the numerator:
f"(x) = [-1 - x + 3 - 3x] / (1+x)^4
10. Combining like terms:
f"(x) = (2x - 2) / (1+x)^4
Thus, the second derivative of f(x) is (2x - 2) / (1+x)^4.