Nigerian Scholars » » If $$y = x \; \mathrm{sin} \; x,$$ find $$\frac{dy}{dx}$$ when $$x = \frac{\pi}{2}.$$...

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) $$\frac{\pi}{2}$$

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

Dicussion (1)

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