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