If y = ( ^ - 1 x), then dy dx is - Vidyadip

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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