Q83 / 150 · MHT CET 2026 (15 April Shift 1) · Mathematics
If $y = \cos^{-1}(\sin x)$, where $\dfrac{\pi}{2} < x < \pi$, then $\dfrac{dy}{dx} = ...$
- A$-1$
- B$0$
- Correct answerC$1$
- D$2$
Solution
Given $y = \cos^{-1}(\sin x)$
Differentiating with respect to $x$ using the chain rule:
$\dfrac{dy}{dx} = \dfrac{-1}{\sqrt{1 - \sin^2 x}} \cdot \dfrac{d}{dx}(\sin x)$
$\dfrac{dy}{dx} = \dfrac{-\cos x}{\sqrt{\cos^2 x}} = \dfrac{-\cos x}{|\cos x|}$
Since $\dfrac{\pi}{2} < x < \pi$, $x$ lies in the second quadrant where $\cos x < 0$.
Therefore, $|\cos x| = -\cos x$.
Substituting this into the derivative:
$\dfrac{dy}{dx} = \dfrac{-\cos x}{-\cos x} = 1$
Answer: $1$
Differentiating with respect to $x$ using the chain rule:
$\dfrac{dy}{dx} = \dfrac{-1}{\sqrt{1 - \sin^2 x}} \cdot \dfrac{d}{dx}(\sin x)$
$\dfrac{dy}{dx} = \dfrac{-\cos x}{\sqrt{\cos^2 x}} = \dfrac{-\cos x}{|\cos x|}$
Since $\dfrac{\pi}{2} < x < \pi$, $x$ lies in the second quadrant where $\cos x < 0$.
Therefore, $|\cos x| = -\cos x$.
Substituting this into the derivative:
$\dfrac{dy}{dx} = \dfrac{-\cos x}{-\cos x} = 1$
Answer: $1$
Hence, the correct answer is option: 'C'.