Question

Dy/dx of ln(x/x^²+1)

Answer 1

$\dfrac{dy}{dx}\ \ln\left(\dfrac{x}{x^2+1}\right)=\dfrac{1}{\frac{x}{x^2+1}}\cdot\dfrac{x'(x^2+1)-x(x^2+1)'}{(x^2+1)^2}\\\\=\dfrac{x^2+1}{x}\cdot\dfrac{1(x^2+1)-x(2x)}{(x^2+1)^2}=\dfrac{1}{x}\cdot\dfrac{x^2+1-2x^2}{x^2+1}=\dfrac{1-x^2}{x^3+x}\\\\Used:\\\\(\ln(x))'=\dfrac{1}{x}\\\\\left[f(g(x))\right]'=f'(g(x))\cdot g'(x)\\\\\left(\dfrac{f(x)}{g(x)}\right)'=\dfrac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}$