Product and Quotient Rule — Question 9

Question 9 - Solution

We are given: $$f(x) = (x^2 + 1)\tan(x)$$

Let: $u(x) = x^2 + 1$ $v(x) = \tan(x)$

We’ll use the product rule: $$f'(x) = u'(x)v(x) + u(x)v'(x)$$

Step 1: Differentiate $u(x)$: $$u'(x) = 2x$$

Step 2: Differentiate $v(x) = \tan(x)$: $$v'(x) = \sec^2(x)$$

Now apply the product rule: $$f'(x) = 2x \tan(x) + (x^2 + 1)\sec^2(x)$$

Final Answer: $$\boxed{ f'(x) = 2x \tan(x) + (x^2 + 1)\sec^2(x) }$$

(b) Where is $f'(x)$ undefined on $(-\pi, \pi)$?

$\tan(x)$ and $\sec^2(x)$ are undefined where $\cos(x) = 0$

$\cos(x) = 0$ at $x = -\frac{\pi}{2}, \frac{\pi}{2}$ in the given interval

So, $f'(x)$ is undefined at: $$\boxed{x = -\frac{\pi}{2}, \; \frac{\pi}{2}}$$