MCQ
If $y = \sin (2{\sin ^{ - 1}}x),$ then ${{dy} \over {dx}} = $
- ✓${{2 - 4{x^2}} \over {\sqrt {1 - {x^2}} }}$
- B${{2 + 4{x^2}} \over {\sqrt {1 - {x^2}} }}$
- C${{2 - 4{x^2}} \over {\sqrt {1 + {x^2}} }}$
- D${{2 + 4{x^2}} \over {\sqrt {1 + {x^2}} }}$
✓
Answer
Correct option: A.
${{2 - 4{x^2}} \over {\sqrt {1 - {x^2}} }}$
a
(a) Let $x = \sin \theta \Rightarrow 2{\sin ^{ - 1}}x = 2\theta $
(a) Let $x = \sin \theta \Rightarrow 2{\sin ^{ - 1}}x = 2\theta $
==> $y = \sin 2\theta $
==>$\frac{{dy}}{{dx}} = \frac{{dy/d\theta }}{{dx/d\theta }} = \frac{{2\cos 2\theta }}{{\cos \theta }}$
$ = \frac{{2(1 - 2{{\sin }^2}\theta )}}{{\sqrt {1 - {{\sin }^2}\theta } }} = \frac{{2 - 4{x^2}}}{{\sqrt {1 - {x^2}} }}$.
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