Calculate f'(0), if f(x) = [ ln(x^2 + 1) ]/(x^2 + 1).

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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

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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

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