MHT CET · Maths · Differentiation
If \(\mathrm{y}=\sin ^{-1}\left[\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right]\), then \(\frac{d y}{d x}=\)
- A \(\left(-\frac{1}{2}\right) \frac{1}{\sqrt{1-x^{2}}}\)
- B \(\left(-\frac{1}{2}\right) \frac{1}{\sqrt{x^{2}-1}}\)
- C \(\left(\frac{1}{4}\right) \frac{1}{\sqrt{x^{2}-1}}\)
- D \(\left(\frac{1}{4}\right) \frac{1}{\sqrt{1-x^{2}}}\)
Answer & Solution
Correct Answer
(A) \(\left(-\frac{1}{2}\right) \frac{1}{\sqrt{1-x^{2}}}\)
Step-by-step Solution
Detailed explanation
The equation of tangent at \(P(-4,-4)\) on the curve \(x^{2}=-4 y\) is
(A) \(3 x-y+8=0\)
(B) \(2 x+y+4=0\)
(C) \(2 x+y-4=0\)
(D) \(2 x-y+4=0\)
(A) \(3 x-y+8=0\)
(B) \(2 x+y+4=0\)
(C) \(2 x+y-4=0\)
(D) \(2 x-y+4=0\)
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