If y = x^0, then dy dx = - Vidyadip

MCQ

If $y = \sec {x^0}$, then ${{dy} \over {dx}} = $

A
$\sec x\tan x$
B
$\sec {x^o}\tan {x^o}$
✓
${\pi \over {180}}\sec {x^o}\tan {x^o}$
D
${{180} \over \pi }\sec {x^o}\tan {x^o}$

✓

Answer

Correct option: C.

${\pi \over {180}}\sec {x^o}\tan {x^o}$

c
(c) As ${x^0} = \frac{{\pi x}}{{180}}$ radian.

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