Q83 / 150 · MHT CET 2026 (19 April Shift 2) · Mathematics

If $y = \sin^{-1}\left(\dfrac{25 - x^2}{25 + x^2}\right)$, then $y'(1)$ is...

  • Correct answer
    A
    $-\dfrac{5}{13}$
  • B
    $\dfrac{5}{13}$
  • C
    $\dfrac{13}{5}$
  • D
    $\dfrac{2}{7}$

Solution

Given $y = \sin^{-1}\left(\dfrac{25 - x^2}{25 + x^2}\right)$

Substitute $x = 5 \tan \theta$, which gives $\theta = \tan^{-1}\left(\dfrac{x}{5}\right)$

$y = \sin^{-1}\left(\dfrac{25 - 25 \tan^2 \theta}{25 + 25 \tan^2 \theta}\right)$

$y = \sin^{-1}\left(\dfrac{1 - \tan^2 \theta}{1 + \tan^2 \theta}\right)$

$y = \sin^{-1}(\cos 2\theta)$

$y = \sin^{-1}\left(\sin\left(\dfrac{\pi}{2} - 2\theta\right)\right)$

For $x = 1$, $\theta = \tan^{-1}\left(\dfrac{1}{5}\right)$, so $2\theta \in \left(0, \dfrac{\pi}{2}\right)$, which lies in the principal domain of $\sin^{-1}$.

$y = \dfrac{\pi}{2} - 2\theta = \dfrac{\pi}{2} - 2 \tan^{-1}\left(\dfrac{x}{5}\right)$

Differentiating with respect to $x$:

$\dfrac{dy}{dx} = -2 \cdot \dfrac{1}{1 + \left(\dfrac{x}{5}\right)^2} \cdot \dfrac{1}{5} = -\dfrac{10}{25 + x^2}$

Substituting $x = 1$:

$y'(1) = -\dfrac{10}{25 + 1} = -\dfrac{10}{26} = -\dfrac{5}{13}$

Answer: $-\dfrac{5}{13}$

Hence, the correct answer is option: 'A'.

Relevant Questions