Question

Find the point on the curve $y=\frac{x}{1+x^2}$, where the tangent to the curve has the greatest slope.

Answer 1

$y=\frac{x}{1+x^2}$

$\frac{dy}{dx}=\frac{(1+x^2)-2x^2}{(1+x^2{)}^2}$

= $\frac{1}{1+x^2}-\frac{2x^2}{(1+x^2{)}^2}=\frac{(1-x{)}^2}{(1+x^2{)}^2}$

$\frac{d^{2}y}{d{x}^2}=\frac{(1+x^2{)}^2[2x-4x]-[(1+x{)}^2-2x^2]\left[2\right(1+x^2.2x]}{(1+x^2{)}^4}$

=$\frac{(1+x^2{)}^2[-2x]+[2x^2(1+x^2)\left]\right[4x(1+x^2)]}{(1+x^2{)}^4}<0$

$\frac{dy}{dx}=0⟹(1-x^2)=0⟹x=1$

$y=\frac{1}{1+1}=\frac{1}{2}$

Therefore, point is $(x,y)=(1,\frac{1}{2})$