COMEDK · Maths · 26. Differentiation
If \(y=\sin ^{-1}\left(\dfrac{1}{\sqrt{x+1}}\right)\) then \(\dfrac{d y}{d x}=\)
- A \(\dfrac{1}{2 x(1+\sqrt{x})}\)
- B \(-\dfrac{1}{2 \sqrt{x}(1+x)}\)
- C \(\dfrac{1}{2 \sqrt{1-x}}\)
- D \(\dfrac{1}{2 \sqrt{x}(1+\sqrt{x})}\)
Answer & Solution
Correct Answer
(B) \(-\dfrac{1}{2 \sqrt{x}(1+x)}\)
Step-by-step Solution
Detailed explanation
Given \(y = \sin^{-1}\left(\dfrac{1}{\sqrt{x+1}}\right)\). Let \(u = \dfrac{1}{\sqrt{x+1}}\). Then \(y = \sin^{-1}(u)\). Using the chain rule, \(\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}\).…
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