27.the slope of tangent to the curve  y= sin^-1(sin x) at x = 3π/4 is
1.1
2.-1
.3.0
4.non-existent
Correct ans is -1 plz exp.
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Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{y}={\sin }^{-1}\left(\mathrm{sinx}\right) \\ \mathrm{Note} \mathrm{that} {\sin }^{-1}\left(\mathrm{sinx}\right)=\mathrm{x} \mathrm{only} \mathrm{for} \mathrm{x}\in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right] \\ \mathrm{Differentiate} \mathrm{y} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x}. \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\sqrt{1-{\sin }^2\mathrm{x}}}.\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{sinx}\right) \\ =\frac{1}{\sqrt{{\cos }^2\mathrm{x}}}.\mathrm{cosx} \\ \mathrm{Now} \mathrm{do} \mathrm{not} \mathrm{write} \sqrt{{\cos }^2\mathrm{x}}=\mathrm{cosx}, \mathrm{because} \mathrm{square} \mathrm{root} \mathrm{function} \mathrm{always} \mathrm{gives} \\ \mathrm{positive} \mathrm{value}, \mathrm{hence} \sqrt{{\cos }^2\mathrm{x}}=\left|\mathrm{cosx}\right| \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\left|\mathrm{cosx}\right|}.\mathrm{cosx} \\ \Rightarrow {\frac{\mathrm{dy}}{\mathrm{dx}}}_{\left(\mathrm{at} \mathrm{x}=\frac{3\mathrm{\pi }}{4}\right)}=\frac{1}{\left|\cos \left(\frac{3\mathrm{\pi }}{4}\right)\right|}.\cos \left(\frac{3\mathrm{\pi }}{4}\right) \\ =\frac{1}{\left|-{\displaystyle \frac{1}{\sqrt{2}}}\right|}\left(-\frac{1}{\sqrt{2}}\right) \\ =\frac{1}{\frac{1}{\sqrt{2}}}.\left(-\frac{1}{\sqrt{2}}\right) \\ =-1 \left(\mathrm{Answer}\right) \end{gathered}$$

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