CalculusHard ShareCalculus · Hard problemFind dydx\dfrac{dy}{dx}dxdy if y=ln (xx2+1)y=\ln\!\left(\dfrac{x}{\sqrt{x^2+1}}\right)y=ln(x2+1x) and simplify to a single fraction.Ax(x2+1)1\dfrac{x(x^2+1)}{1}1x(x2+1)B1x(x3+1)\dfrac{1}{x(x^{3}+1)}x(x3+1)1C1x(x2+2)\dfrac{1}{x(x^2+2)}x(x2+2)1D1x(x2+1)\dfrac{1}{x(x^2+1)}x(x2+1)1Check Show answer Walk through it step by step Practice more like this