A) \[\frac{1}{\sqrt{1-{{x}^{2}}}}\]
B) \[\frac{-1}{\sqrt{1-{{x}^{2}}}}\]
C) \[\frac{2}{\sqrt{1-{{x}^{2}}}}\]
D) \[\frac{-2}{\sqrt{1-{{x}^{2}}}}\]
Correct Answer: B
Solution :
[b] putting \[x=\cos \theta \] \[\therefore y=soi{{n}^{-1}}\left\{ \frac{5.\cos \theta +12.\sin \theta }{13} \right\}={{\sin }^{-1}}\left[ \left( \frac{5}{13} \right).\cos \theta +\frac{12}{13} \right].\sin \theta \]Let \[\frac{5}{13}=\sin \alpha ,\] then\[\Rightarrow \frac{12}{13}=\cos \alpha ,\] \[\therefore y={{\sin }^{-1}}\{\sin (\alpha +\theta )\}=\alpha +\theta \] Differentiating w.r.t. x, we have \[\frac{dy}{d\theta }=\frac{d\theta }{dx}=\frac{d}{dx}({{\cos }^{-1}}x)=\frac{-1}{\sqrt{1-{{x}^{2}}}}\] Hence, option [b] is correct.
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