Let f(x)= ln (1/x^2 + 1) find f'(x)....can anyone help with this calculus problem?

Question

Finding the derivative...do you just take the deriv..1/1/x^2 + 1 (2x)..i don't know..please help if you know...thanks a lot!

Answer 1

this is the derivative of a composite fcn

i am assuming the denominator to be (x^2) + 1 so that
let u = 1/(x^2 + 1)

then du/dx = (-1)(2x)(x^2 + 1)^(-2) = (-2x)/(x^2 + 1)^2

y = Lnu

dy/du = 1/u

dy/dx = (dy/du)(du/dx) = (1/u)(-2x)/(x^2 + 1)^2

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

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

Answer 2

f´(x) =(x^2+1)* 1/(x^2+1)^2 *(-2x) = (-2x)/(x^2+1)

I used f(x) = ln [1/(x^2+1)]

Answer 3

f(x)= ln (1/x^2 + 1)
this can be written in a simpler form
f(x)= ln (1) -ln(x^2 +1) or -ln(x^2+1)
d [log u]/dx = (1/u)*du/dx u=x^2 +1 du=2x
f'(x)= -2x/x^2 +1
good luck

Answer 4

it can be broken into
ln(z)
z = 1/(x^2 +1)
z'= (-1/(x^2+1)^2)*2x
ln(z) = (1/z)*z'
= -2x(x^2+1)/(x^2+1)^2
= -2x/(x^2+1)