Find the derivative of this function at $x_o = \pi$

Question

Find the derivative of $g(x)= (\tan \left|x\right| + x )\sin(x) $ at $x_o = \pi$.

I tried to solve it using $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$, but I got stuck at this point:

$\lim_{h \to 0} \ \frac {\tan \left|\pi+ h\right| + \pi+h )\sin(\pi+h)-\tan \left|\pi\right| + \pi )\sin(\pi)}{h}$.

Since $h \to 0$, it seems to me that the derivative is $0$, but I have no idea how to prove it.

Answer 1

Using the product rule:

$$g'(x) = (\tan^2 |x| +1) \sin x + \cos x \tan|x| + \sin x + x\cos x$$ $$g'(x) = \tan^2 |x| \sin x +\sin x + \cos x \tan|x| + \sin x + x\cos x$$ at $x_0= \pi$

$$g'(\pi) = -\pi$$

Answer 2

We have that

$$\lim_{h \to 0} \ \frac {(\tan \left|\pi+ h\right| + \pi+h )\sin(\pi+h)-\tan \left|\pi\right| + \pi )\sin(\pi)}{h}=$$ $$=\lim_{h \to 0} \ \frac {(\tan \left|\pi+ h\right| + \pi+h )\sin(\pi+h)}{h}=$$ $$=\lim_{h \to 0} \ (\tan \left|\pi+ h\right| + \pi+h )\frac{-\sin h}{h}=\pi \cdot (-1)=-\pi$$

Answer 3

Since you're computing the derivative at $\pi$, you can safely disregard the absolute value, because $x$ is positive in a neighborhood of $\pi$, so in such a neighborhood of $\pi$ you have $\tan\lvert x\rvert=\tan x$.

The product rule now tells you $$ f'(\pi)=(1+\tan^2\pi+1)\sin\pi+(\tan\pi+\pi)\cos\pi=-\pi $$

It would be different if the derivative at $0$ is to be computed.