Differentiation Formulas — Question 1

Question 1 - Solution

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

(a) Find $f'(x)$:

Differentiate term by term.

First term (quotient rule): $$\frac{d}{dx} \left( \frac{\sin(x)}{x^2 + 1} \right) = \frac{(x^2 + 1)\cos(x) - \sin(x)(2x)}{(x^2 + 1)^2}$$

Second term (product rule): $$\frac{d}{dx}(x \cos(x)) = \cos(x) - x\sin(x)$$

Putting both together: $$f'(x) = \frac{(x^2 + 1)\cos(x) - 2x\sin(x)}{(x^2 + 1)^2} + \cos(x) - x\sin(x)$$

(b) Find the slope at $x = 0$:

Evaluate $f'(0)$.

$$\begin{align*} f'(0) &= \frac{(0^2 + 1)\cos(0) - 2(0)\sin(0)}{(0^2 + 1)^2} + \cos(0) - 0\cdot\sin(0) \\ &= \frac{1\cdot 1 - 0}{1^2} + 1 - 0 \\ &= 1 + 1 = \boxed{2} \end{align*}$$

So, the slope of the tangent line to the curve at $x = 0$ is $\boxed{2}$.