Question

Q46


$$\begin{gathered} \mathbf{46}\mathbf{.}Ify=\frac{\sqrt{1-{\sin }^{-1}x}}{1+{\sin }^{-1}x}then{y}^1\left(0\right)= \\ a)0b)1 \\ c)-1d)noneofthese \end{gathered}$$

Answer 1

Dear student
$$\begin{gathered} \mathrm{y}=\frac{\sqrt{1-{\sin }^{-1}\mathrm{x}}}{1+{\sin }^{-1}\mathrm{x}} \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}\mathrm{x}\mathrm{we}\mathrm{get} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left[\sqrt{1-{\sin }^{-1}\mathrm{x}}\right]\left(1+{\sin }^{-1}\mathrm{x}\right)-\sqrt{1-{\sin }^{-1}\mathrm{x}}{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left[1+{\sin }^{-1}\mathrm{x}\right]}{{\left(1+{\sin }^{-1}\mathrm{x}\right)}^2} \\ =\frac{{\displaystyle \frac{1}{2}}{\left(1-{\sin }^{-1}\mathrm{x}\right)}^{{\displaystyle \frac{1}{2}}-1}{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(1-{\sin }^{-1}\mathrm{x}\right)\left(1+{\sin }^{-1}\mathrm{x}\right)-\left({\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left[{\sin }^{-1}\mathrm{x}\right]+{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(1\right)\right)\sqrt{1-{\sin }^{-1}\mathrm{x}}}{{\left(1+{\sin }^{-1}\mathrm{x}\right)}^2} \\ =\frac{{\displaystyle \frac{\left\{{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(1\right)-{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left({\sin }^{-1}\mathrm{x}\right)\right\}\left[1+{\sin }^{-1}\mathrm{x}\right]}{2\sqrt{1-{\sin }^{-1}\mathrm{x}}}}-\left({\displaystyle \frac{1}{\sqrt{1-{\mathrm{x}}^2}}}+0\right)\sqrt{1-{\sin }^{-1}\mathrm{x}}}{{\left(1+{\sin }^{-1}\mathrm{x}\right)}^2} \\ =\frac{-{\displaystyle \frac{1+{\sin }^{-1}\mathrm{x}}{2\sqrt{1-{\mathrm{x}}^2}\sqrt{1-{\sin }^{-1}\mathrm{x}}}}-{\displaystyle \frac{\sqrt{1-{\sin }^{-1}\mathrm{x}}}{\sqrt{1-{\mathrm{x}}^2}}}}{{\left(1+{\sin }^{-1}\mathrm{x}\right)}^2} \\ \mathrm{So},\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-{\displaystyle \frac{1+{\sin }^{-1}\mathrm{x}}{2\sqrt{1-{\mathrm{x}}^2}\sqrt{1-{\sin }^{-1}\mathrm{x}}}}-{\displaystyle \frac{\sqrt{1-{\sin }^{-1}\mathrm{x}}}{\sqrt{1-{\mathrm{x}}^2}}}}{{\left(1+{\sin }^{-1}\mathrm{x}\right)}^2} \\ \mathrm{Now},\mathrm{y}\text{'}\left(0\right)=\frac{-{\displaystyle \frac{1+{\sin }^{-1}\left(0\right)}{2\sqrt{1-(0{)}^2}\sqrt{1-{\sin }^{-1}\left(0\right)}}}-{\displaystyle \frac{\sqrt{1-{\sin }^{-1}\left(0\right)}}{\sqrt{1-(0{)}^2}}}}{{\left(1+{\sin }^{-1}\left(0\right)\right)}^2} \\ =\frac{-{\displaystyle \frac{1+0}{2\sqrt{1-0}\sqrt{1-0}}}-{\displaystyle \frac{\sqrt{1-0)}}{\sqrt{1-0}}}}{{\left(1+0\right)}^2} \\ =\frac{-{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{1}}}{1} \\ =-\frac{3}{2} \end{gathered}$$
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