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प्रश्न
Mark the correct alternative in of the following:
If f(x) = x sinx, then \[f'\left( \frac{\pi}{2} \right) =\]
पर्याय
0
1
−1
\[\frac{1}{2}\]
उत्तर
f(x) = x sinx
Differentiating both sides with respect to x, we get
\[f'\left( x \right) = x \times \frac{d}{dx}\left( \sin x \right) + \sin x \times \frac{d}{dx}\left( x \right) \left( \text{ Product rule } \right)\]
\[ = x \times \cos x + \sin x \times 1\]
\[ = x \cos x + \sin x\]
Putting \[x = \frac{\pi}{2}\]
we get \[f'\left( \frac{\pi}{2} \right) = \frac{\pi}{2} \times \cos\left( \frac{\pi}{2} \right) + \sin\left( \frac{\pi}{2} \right)\]
\[ = \frac{\pi}{2} \times 0 + 1\]
\[ = 1\]
Hence, the correct answer is option (b).
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