Find $f'(8.23)$ where $f(x)=23|x|−37\lfloor x\rfloor+58\{x\}+88\arccos(\sin x)−40\max(x,0)$

Question

Let $$f(x)=23|x|−37\lfloor x\rfloor+58\{x\}+88\arccos(\sin x)−40\max(x,0).$$ Find $f^\prime(8.23)$. Note: For a real number $x$, $\{x\}=x−\lfloor x\rfloor$ denotes the fractional part of x. I don't know the derivatives of the few pieces of this function (like the fractional part).

Answer 1

Let us examine this function near $8.23$, for example, on the interval $(8,9)$. Now, since $x\in(8,9)$, we can say the following:

1. $|x|=x$
2. $\lfloor x\rfloor=8$
3. $\{x\}=x-\lfloor x\rfloor=x-8$
4. $\max(x,0)=x$

The function then simplifies to

$$\begin{align}f(x)&=23x−37(8)+58(x-8)+88\arccos(\sin(x))−40x\\ &=(23+58-40)x+88\arccos(\sin(x)) - 8(58+37)\\ &=41x+88\arccos(\sin(x)) - 760\\ \end{align}$$

Then, we can differentiate

$$\begin{align}f'(x)&=41+88\left(-\frac{1}{\sqrt{1-\sin^2(x)}}\cos(x)\right)\\ &=41+88\left(-\frac{\cos(x)}{|\cos(x)|}\right)\\ \end{align}$$

All that is left is to find the sign of $\cos(8.23)$. $8.23>2.5\cdot3.15>2.5\pi$ and $8.23<3\cdot3.14<3\pi$, so the sign is negative:

$$f'(8.23)=41+88=129$$

Answer 2

If you use the fact, that the derivative only depends on a small neighbourhood around the point ($8.23$ in this case) you can represent the function by a much simpler expression in that region.

For example if you choose the neighbourhood small enough around $8.23$, all the points in it will be positive. Then $|x| = x$ in this neighbourhood.