MCQ
If $y = {\cot ^{ - 1}}({x^2})$, then ${{dy} \over {dx}}$ is equal to
- A${{2x} \over {1 + {x^4}}}$
- B${{2x} \over {\sqrt {1 + 4x} }}$
- ✓${{ - 2x} \over {1 + {x^4}}}$
- D${{ - 2x} \over {\sqrt {1 + {x^2}} }}$
✓
Answer
Correct option: C.
${{ - 2x} \over {1 + {x^4}}}$
c
(c) $y = {\cot ^{ - 1}}\left( {{x^2}} \right)$
(c) $y = {\cot ^{ - 1}}\left( {{x^2}} \right)$
$\frac{{dy}}{{dx}} = \frac{{ - 1}}{{1 + {{({x^2})}^2}}}\frac{d}{{dx}}({x^2}) $
$= \frac{{ - 1}}{{1 + {x^4}}}(2x) = \frac{{ - 2x}}{{1 + {x^4}}}$.
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