MCQ
If $y = \sec ({\tan ^{ - 1}}x),$ then ${{dy} \over {dx}}$ is
  • ${x \over {\sqrt {1 + {x^2}} }}$
  • B
    ${{ - x} \over {\sqrt {1 + {x^2}} }}$
  • C
    ${x \over {\sqrt {1 - {x^2}} }}$
  • D
    None of these

Answer

Correct option: A.
${x \over {\sqrt {1 + {x^2}} }}$
a
(a) $y = \sec ({\tan ^{ - 1}}x) {{dy} \over {dx}}$

$= \sec ({\tan ^{ - 1}}x) \tan ({\tan ^{ - 1}}x) \cdot \frac{{1}}{{1 + {x^2}}}$

$ = \frac{{x}}{{1 + {x^2}}}\,.\,\sqrt {1 + {x^2}} = \frac{{x}}{{\sqrt {1 + {x^2}} }}$,   $({\tan ^{ - 1}}x = {\sec ^{ - 1}}\sqrt {1 + {x^2}} )$.

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