A) 1
B) \[\frac{1}{1+{{x}^{2}}}\]
C) 2
D) None of these
Correct Answer: A
Solution :
Let\[{{y}_{1}}={{\sin }^{-1}}x\] and \[{{y}_{2}}={{\cos }^{-1}}\sqrt{1-{{x}^{2}}}\] Differentiating w.r.t. x of \[{{y}_{1}}\]and \[{{y}_{2}}\], we get \[\frac{d{{y}_{1}}}{dx}=\frac{1}{\sqrt{1-{{x}^{2}}}}\] \[\frac{d{{y}_{2}}}{dx}=-\frac{1}{\sqrt{1-(1-{{x}^{2}})}}\frac{1(-2x)}{2\sqrt{1-x}}=\frac{1}{\sqrt{1-{{x}^{2}}}}\Rightarrow \frac{d{{y}_{2}}}{d{{y}_{1}}}=1.\]
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