Mathematics - Application of Derivative Question with Solution | TestHub

$y=1\Rightarrow x=\pm 1$; points are $(1,1),(-1,1);\frac{dy}{dx}=\frac{16x}{{\left(3+x^2\right)}^2},{\left(\frac{dy}{dx}\right)}_{(1,1)}=1,{\left(\frac{dy}{dx}\right)}_{(-1,1)}=-1$ Eq. of tangent at $(1,1)$ is $y-1=(x-1)\Rightarrow x-y=0$ Eq. of tangent at $(-1,1):y-1=-1(x+1)\Rightarrow x+y=0$ Both tangents pass through origin.