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5. continuity and differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths5. continuity and differentiation1 Mark
MCQ
If $y = {\sin ^{ - 1}}\sqrt {1 - {x^2}} $, then $dy/dx = $
  • A
    ${1 \over {\sqrt {1 - {x^2}} }}$
  • B
    ${1 \over {\sqrt {1 + {x^2}} }}$
  • ✓
    $ - {1 \over {\sqrt {1 - {x^2}} }}$
  • D
    $ - {1 \over {\sqrt {{x^2} - 1} }}$
✓

Answer

Correct option: C.
$ - {1 \over {\sqrt {1 - {x^2}} }}$
c
(c) $y = {\sin ^{ - 1}}(\sqrt {1 - {x^2}} )$

Let $\sqrt {1 - {x^2}} = \sin \theta \Rightarrow 1 - {x^2} = {\sin ^2}\theta $

==> ${x^2} = 1 - {\sin ^2}\theta = {\cos ^2}\theta $

$\therefore x = \cos \theta $ or $\theta = {\cos ^{ - 1}}x$ 

$ \Rightarrow y = {\cos ^{ - 1}}x$

Differentiating w.r.t.  $x $ of $y,$  we get 

$\frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {1 - {x^2}} }}$.

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