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STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
MCQ
${d \over {dx}}{\tan ^{ - 1}}{x \over {\sqrt {{a^2} - {x^2}} }} = $
  • A
    ${a \over {{a^2} + {x^2}}}$
  • B
    ${{ - a} \over {{a^2} + {x^2}}}$
  • C
    ${1 \over {a\sqrt {{a^2} - {x^2}} }}$
  • ✓
    ${1 \over {\sqrt {{a^2} - {x^2}} }}$
✓

Answer

Correct option: D.
${1 \over {\sqrt {{a^2} - {x^2}} }}$
d
(d) $\frac{d}{{dx}}{\tan ^{ - 1}}\frac{x}{{\sqrt {{a^2} - {x^2}} }}$

Putting $x = a\sin \theta ,$ we get

$ = \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\frac{{a\sin \theta }}{{a\cos \theta }}} \right] = \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\tan \theta } \right] = \frac{d}{{dx}}[\theta ]$

Substituting value of $\theta $,  so

$= \frac{d}{dx}\left[ {{\sin }^{-1}}\tan \left( \frac{x}{a} \right) \right]=\frac{1}{\sqrt{{{a}^{2}}-{{x}^{2}}}}.$

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