d dx ( x^2 1 x ) = - Vidyadip

MCQ

${d \over {dx}}\left( {{x^2}\sin {1 \over x}} \right) = $

A
$\cos \,\left( {{1 \over x}} \right) + 2x\sin \left( {{1 \over x}} \right)$
✓
$2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right)$
C
$\cos \left( {{1 \over x}} \right) - 2x\sin \left( {{1 \over x}} \right)$
D
None of these

✓

Answer

Correct option: B.

$2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right)$

b
(b) $\frac{d}{{dx}}\left( {{x^2}\sin \frac{1}{x}} \right) = {x^2}\cos \left( {\frac{1}{x}} \right)\frac{d}{{dx}}\left( {\frac{1}{x}} \right)$$ + 2x\sin \left( {\frac{1}{x}} \right)$

$ = - \frac{1}{{{x^2}}}.{x^2}\cos \left( {\frac{1}{x}} \right) + 2x\sin \left( {\frac{1}{x}} \right) = 2x\sin \left( {\frac{1}{x}} \right) - \cos \left( {\frac{1}{x}} \right)$.

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