A) \[-\frac{1}{2x\sqrt{1-x}}-\frac{1}{\sqrt{1-{{x}^{2}}}}\]
B) \[\frac{1}{2\sqrt{x}\sqrt{1-x}}-\frac{1}{\sqrt{1-{{x}^{2}}}}\]
C) \[\frac{1}{2\sqrt{x}\sqrt{1-x}}+\frac{1}{\sqrt{1-{{x}^{2}}}}\]
D) \[-\frac{1}{2\sqrt{x}\sqrt{1-x}}+\frac{1}{\sqrt{1-{{x}^{2}}}}\]
Correct Answer: C
Solution :
\[\frac{d}{dx}\{{{\sin }^{-1}}(x\sqrt{1-x}+\sqrt{x}\sqrt{1-{{x}^{2}}})\}\] \[=\frac{d}{dx}{{\sin }^{-1}}\{(x\sqrt{1-{{(\sqrt{x})}^{2}}}+\sqrt{x}\sqrt{1-{{x}^{2}}})\}\] \[\left[ \begin{align} & \because \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\sin }^{-1}}A+{{\sin }^{-1}}B \\ & ={{\sin }^{-1}}(A\sqrt{1-{{B}^{2}}}+B\sqrt{1-{{A}^{2}}}) \\ \end{align} \right]\] \[\Rightarrow \] \[\frac{d}{dx}\{{{\sin }^{-1}}(x\sqrt{1-{{(\sqrt{x})}^{2}}}+\sqrt{x}\sqrt{1-{{x}^{2}}})\}\] \[=\frac{d}{dx}({{\sin }^{-1}}x+{{\sin }^{-1}}\sqrt{x})\] \[=\frac{1}{\sqrt{1-{{x}^{2}}}}+\frac{1}{2\sqrt{x}\sqrt{1-x}}\]
You need to login to perform this action.
You will be redirected in
3 sec
