The coordinates of the point on the curve $y=\frac{x}{1+x^{2}}$ where the tangent to the curve has t

Question

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

Professor PC Thomas Classes · Answer

$y=\frac{x}{1+x^{2}}$\ $\frac{d y}{d x}=\frac{\left(1+x^{2}ight) \cdot 1-x(2 x)}{\left(1+x^{2}ight)^{2}}=\frac{1-x^{2}}{\left(1+x^{2}ight)^{2}}$\ Slope of the tangent $=m=\frac{d y}{d x}=\frac{1-x^{2}}{\left(1+x^{2}ight)^{2}}$ $$ \begin{aligned} \frac{1 n}{d x} & =\frac{\left(1+x^{2}ight)^{2}(-2 x)-\left(1-x^{2}ight) \cdot 2\left(1+x^{2}ight) 2 x}{\left(1+x^{2}ight)^{4}} \ & =\frac{2 x^{3}-6 x}{\left(1+x^{2}ight)^{3}} \end{aligned} $$ For greatest or least values of $m$ we should have\ $\frac{d m}{d x}=0 \Rightarrow x=0, x= \pm \sqrt{3}$\ Now, $\frac{d^{2} m}{d x^{2}}$\ $=\frac{\left(1+x^{2}ight)^{3}\left(6 x^{2}-6ight)-\left(2 x^{3}-6 xight) \cdot 3\left(1+x^{2}ight)^{2} \cdot 2 x}{\left(1+x^{2}ight)^{6}}$\ $\left(\frac{d^{2} m}{d x^{2}}ight)_{	ext {at } x=0}=-6<0$\ $	herefore \quad m$ has a maximum value at $x=0$ and when $x=0, y=0$\ $	herefore \quad$ The tangent to the curve has maximum slope at $(0,0)$.\