CUET Preparation Today
If $y=\tan ^{-1} x$ then $\frac{dy}{dx}$ at $y=\frac{\pi}{3}$ is equal to: |
1 $\frac{1}{2}$ $\frac{1}{3}$ $\frac{1}{4}$ |
$\frac{1}{4}$ |
The correct answer is Option (4) → $\frac{1}{4}$ $\frac{dy}{dx}=\frac{1}{1+x^2}⇒\left.\frac{dy}{dx}\right]_{at\,y=\frac{\pi}{3}}$ $\tan^{-1}x=\frac{\pi}{3}⇒x=\sqrt{3}$ $\frac{dy}{dx}=\frac{1}{1+{\sqrt{3}}^2}$ $=\frac{1}{4}$ |
