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If \(y = x \; \mathrm{sin} \; x,\) find \(\frac{dy}{dx}\) when \(x = \frac{\pi}{2}.\)...


Question

If \(y = x \; \mathrm{sin} \; x,\) find \(\frac{dy}{dx}\) when \(x = \frac{\pi}{2}.\)

Options

A) \(\frac{- \pi}{2}\)

B) \(-1\)

C) \(1\)

D) \(\frac{\pi}{2}\)

The correct answer is C.

Explanation:

\(y = x \; \mathrm{sin} \; x\)
\(\frac{dy}{dx} = 1 \times \mathrm{sin} \; x + x \; \mathrm{cos} \; x\)
\(= \mathrm{sin} \; x + x \; \mathrm{cos} \; x\)
At \(x = \frac{\pi}{2},\) \(= \mathrm{sin} \; \frac{\pi}{2} + \frac{\pi}{2} \; \mathrm{cos} \frac{π}{2}\)
\(= 1 + \frac{\pi}{2} \times (0) = 1\)


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Discussion (3)

  • Sunday

    The answer: -pi/2
    because Π÷2 is =180
    and cos 180=_-1

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    1. Proffyemphy

      π/2 Is not 180 it's 90

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  • \(y = x \; \mathrm{sin} \; x\)
    \(\frac{dy}{dx} = 1 \times \mathrm{sin} \; x + x \; \mathrm{cos} \; x\)
    \(= \mathrm{sin} \; x + x \; \mathrm{cos} \; x\)
    At \(x = \frac{\pi}{2},\) \(= \mathrm{sin} \; \frac{\pi}{2} + \frac{\pi}{2} \; \mathrm{cos} \frac{π}{2}\)
    \(= 1 + \frac{\pi}{2} \times (0) = 1\)