If \(y = x\tan^{-1}x\), then the value of \(\displaystyle\frac{dy}{dx}\) at \(x = 1\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{\pi}{4} + \frac{1}{2}\)
By the product rule, using \(\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^{2}}\):
\( \displaystyle \frac{dy}{dx} = 1 \cdot \tan^{-1}x + x \cdot \frac{1}{1 + x^{2}} \)
At x = 1, \(\tan^{-1}1 = \frac{\pi}{4}\):
\( \displaystyle \frac{dy}{dx} = \frac{\pi}{4} + \frac{1}{1 + 1} \)
\( \displaystyle = \frac{\pi}{4} + \frac{1}{2} \)