the second derivative of f(x)=ln(x^2+1)
f' = 2 x /(x^2+1)
f" = [ 2(x^2+1)- 2x(2x) ]/(x^2+1)^2
= [ 2x^2+2 -4 x^2 ]/(x^2+1)^2
2[1-x^2] /(x^2+1)^2
can't we cancel them out
cancel what out? There are no common factors.
To find the second derivative of the function f(x) = ln(x^2 + 1), we'll need to find the first derivative first, and then differentiate it again. Let's do it step by step:
Step 1: Find the first derivative of f(x):
To differentiate f(x), we can use the chain rule. The chain rule states that for a composite function, such as ln(g(x)), the derivative is given by g'(x)/g(x). Applying the chain rule to our function, we get:
f'(x) = (1 / (x^2 + 1)) * (2x)
= 2x / (x^2 + 1)
So, the first derivative of f(x) is 2x / (x^2 + 1).
Step 2: Find the second derivative of f(x):
To find the second derivative, we differentiate the first derivative we found in Step 1. Using the quotient rule this time, which states that if we have a division of two functions, (u/v), where u and v are functions of x, then the derivative is given by:
(u'v - uv') / v^2
Let's differentiate the first derivative we found:
f''(x) = [(2 * (x^2 + 1)) - (2x * 2x)] / (x^2 + 1)^2
= (2x^2 + 2) - (4x^2) / (x^2 + 1)^2
= -2x^2 + 2 / (x^2 + 1)^2
Therefore, the second derivative of f(x) = ln(x^2 + 1) is -2x^2 + 2 / (x^2 + 1)^2.