Product and Quotient Rule — Question 7

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Question 7

Let f(x)=xsin⁡xx2+1f(x) = \frac{x \sin x}{x^2 + 1}

  • (a) Compute f′(x)f'(x) using the quotient rule.

  • (b) Simplify the derivative fully.

Original worksheet page 1: question and worked solution for 3-4-007
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Question 7 - Solution

We are given: f(x)=xsin⁡xx2+1f(x) = \frac{x \sin x}{x^2 + 1}

Let: u(x)=xsin⁡x,v(x)=x2+1u(x) = x \sin x, \quad v(x) = x^2 + 1

We apply the quotient rule: f′(x)=u′(x)v(x)−u(x)v′(x)[v(x)]2f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}

First, compute u′(x)u'(x) using the product rule: u′(x)=sin⁡x+xcos⁡xu'(x) = \sin x + x \cos x

Next, compute v′(x)v'(x): v′(x)=2xv'(x) = 2x

Now apply the quotient rule: f′(x)=(sin⁡x+xcos⁡x)(x2+1)−xsin⁡x(2x)(x2+1)2f'(x) = \frac{(\sin x + x \cos x)(x^2 + 1) - x \sin x (2x)}{(x^2 + 1)^2}

Expand the numerator: (sin⁡x)(x2+1)+xcos⁡x(x2+1)−2x2sin⁡x(\sin x)(x^2 + 1) + x \cos x (x^2 + 1) - 2x^2 \sin x

Distribute: x2sin⁡x+sin⁡x+x3cos⁡x+xcos⁡x−2x2sin⁡xx^2 \sin x + \sin x + x^3 \cos x + x \cos x - 2x^2 \sin x

Combine like terms: −x2sin⁡x+sin⁡x+x3cos⁡x+xcos⁡x- x^2 \sin x + \sin x + x^3 \cos x + x \cos x

Factor where possible: sin⁡x(1−x2)+xcos⁡x(x2+1)\sin x (1 - x^2) + x \cos x (x^2 + 1)

Final Answer: f′(x)=sin⁡x(1−x2)+xcos⁡x(x2+1)(x2+1)2\boxed{ f'(x) = \frac{ \sin x (1 - x^2) + x \cos x (x^2 + 1) }{(x^2 + 1)^2} }

Original worksheet page 2: question and worked solution for 3-4-007

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