How do you find the derivative of ln(x/(x^2+1))?

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##f'(x)=(1-x^2)/(x^3+x)##

The easiest way is to primary rewrite the business using properties of logarithms. Recall that ##log(a/b)=log(a)-log(b)##.

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

##f(x)=ln(x)-ln(x^2+1)##

Now we feel two simpler businesss to individualize. Recall that ##d/dxln(x)=1/x##. The tells us that the derivative of a business amid ##ln(x)## is ##d/dxln(g(x))=1/(g(x))*g'(x)=(g'(x))/(g(x))##.

Then:

##f'(x)=1/x-(d/dx(x^2+1))/(x^2+1)##

##f'(x)=1/x-(2x)/(x^2+1)##

These can be wholly, but it's not necessary:

##f'(x)=(x^2+1-2x(x))/(x(x^2+1))##

##f'(x)=(1-x^2)/(x^3+x)##

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