d dx [ 2 x^0 ] = - Vidyadip

MCQ

${d \over {dx}}\left[ {{2 \over \pi }\sin {x^0}} \right] = $

A
${\pi \over {180}}\cos {x^0}$
✓
${1 \over {90}}\cos {x^0}$
C
${\pi \over {90}}\cos {x^0}$
D
${2 \over {90}}\cos {x^0}$

✓

Answer

Correct option: B.

${1 \over {90}}\cos {x^0}$

b
(b) $\frac{d}{{dx}}\left[ {\frac{2}{\pi }\sin x^\circ } \right] = \frac{d}{{dx}}\left[ {\frac{2}{\pi }\sin \frac{{\pi .x}}{{180}}} \right]$

$ = \frac{2}{\pi }\frac{\pi }{{180}}\cos \frac{{x\pi }}{{180}} = \frac{{\cos x^\circ }}{{90}}$ .

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