JAMB 2017 Mathematics · Question 29

If y = x Sin x, find dydx\frac{dy}{dx} when x = π2\frac{\pi}{2}

Options

  1. A.−π2\frac{- \pi}{2}
  2. B.-1
  3. C.1
  4. D.π2\frac{ \pi}{2}

Answer: C. 1

Explanation written with AI and checked against the answer key

Analysis

To solve this problem, we need to find the derivative of the function y=xsin⁡xy = x \sin x with respect to xx. Since yy is the product of two functions of xx (xx and sin⁡x\sin x), we must use the Product Rule of differentiation.

Concept: The Product Rule

The product rule states that if y=u⋅vy = u \cdot v, where uu and vv are functions of xx, then:

dydx=udvdx+vdudx\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}

In this case:

  • u=xu = x
  • v=sin⁡xv = \sin x
Step-by-Step Solution

Step 1: Differentiate the individual parts.

  • dudx=ddx(x)=1\frac{du}{dx} = \frac{d}{dx}(x) = 1
  • dvdx=ddx(sin⁡x)=cos⁡x\frac{dv}{dx} = \frac{d}{dx}(\sin x) = \cos x

Step 2: Apply the Product Rule formula.

dydx=(x)(cos⁡x)+(sin⁡x)(1)\frac{dy}{dx} = (x)(\cos x) + (\sin x)(1)
dydx=xcos⁡x+sin⁡x\frac{dy}{dx} = x \cos x + \sin x

Step 3: Substitute x=π2x = \frac{\pi}{2} into the derivative.

dydx at x=π2=(π2)cos⁡(π2)+sin⁡(π2)\frac{dy}{dx} \text{ at } x = \frac{\pi}{2} = \left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right)

Step 4: Evaluate the trigonometric functions.

  • Recall that cos⁡(π2)=0\cos(\frac{\pi}{2}) = 0
  • Recall that sin⁡(π2)=1\sin(\frac{\pi}{2}) = 1

Substituting these values:

dydx=(π2⋅0)+1\frac{dy}{dx} = \left(\frac{\pi}{2} \cdot 0\right) + 1
dydx=0+1\frac{dy}{dx} = 0 + 1
dydx=1\frac{dy}{dx} = 1
Conclusion

The value of the derivative at x=π2x = \frac{\pi}{2} is 1. This matches option C.

Exam Tip

When dealing with trigonometric functions in calculus, always ensure your calculator is in Radian mode if you are calculating values, though knowing your unit circle values (sin⁡90∘=1,cos⁡90∘=0\sin 90^\circ=1, \cos 90^\circ=0) is much faster and prevents errors during JAMB exams!

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