f(x) = ln(x^2+1)/(x^2+1)

Th function f(x) is a product of two functions.

Let us assume f(x)=u/v, where:

u= ln(x^2+1) ==> u'= 2x(1/(x^2+1)

v= x^2+1 ==> v'= 2x

Now f'(x) = (u'v-uv')/v^2

= [2x(1/(x^2+1)*(x^2+1) -ln(x^2+1)*2x]/(x^2+1)^2

= 2x-2xln(x^2+1)]/(x^2+1)^2

= 2x(1-ln(x^2+1))/(x^2+1)^2

f'(0) = 0/1 = 0

In order to calculate the value of the derivative f'(0), we'll have to find first, the derivative of the function.

We'll use the quotient law, so that:

f'(x) = {[ln(x^2 + 1)]'*(x^2 + 1) - [ln(x^2 + 1)]*(x^2 + 1)'}/(x^2 + 1)^2

f'(x)={2x(x^2 + 1)/(x^2 + 1)-2x*[ln(x^2 + 1)]}/(x^2 + 1)^2

After reducing similar terms and factorizing, we'll get:

f'(x)=2x*[1-ln(x^2 + 1)]/(x^2 + 1)^2

Now, we'll calculate f'(0), substituting x by 0, in the expression of f'(x).

f'(0) = 0*[1-ln(0 + 1)]/(0 + 1)^2

f'(0) = 0/1 = 0

Another way to calculate f'(0), would be to use the definition of the derivative of the function:

f'(0) = lim [f(x)-f(0)]/(x-0)

To find f'(x), if f(x) = ln(x^2+1)]/(x^2+1)

Solution:

Let y= lnt/t^2, where t = x^2+1.So that dt/dx = 2x.

Therefore, dy/dx = f'(x) = [(lnt)'/t - lnt/t^2] (2x)

= { (1/(x^2+1)^2 - ln(x^2+1)/(x^2+1) }(2x)

Threfore f'(0) = {...}(2*0) = 0.