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