Product and Quotient Rule — Question 7

Question 7

Let $$f(x) = \frac{x \sin x}{x^2 + 1}$$

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

• Simplify the derivative fully.

Question 7 - Solution

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

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

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

First, compute $u'(x)$ using the product rule: $$u'(x) = \sin x + x \cos x$$

Next, compute $v'(x)$: $$v'(x) = 2x$$

Now apply the quotient rule: $$f'(x) = \frac{(\sin x + x \cos x)(x^2 + 1) - x \sin x (2x)}{(x^2 + 1)^2}$$

Expand the numerator: $$(\sin x)(x^2 + 1) + x \cos x (x^2 + 1) - 2x^2 \sin x$$

Distribute: $$x^2 \sin x + \sin x + x^3 \cos x + x \cos x - 2x^2 \sin x$$

Combine like terms: $$- x^2 \sin x + \sin x + x^3 \cos x + x \cos x$$

Factor where possible: $$\sin x (1 - x^2) + x \cos x (x^2 + 1)$$

Final Answer: $$\boxed{ f'(x) = \frac{ \sin x (1 - x^2) + x \cos x (x^2 + 1) }{(x^2 + 1)^2} }$$