Determining derivatives of trigonometric functions For what function $f$ and number $a$ is the limit
$$\lim_{x \to \pi/4} \frac {\tan x - 1}{4x-\pi}$$
the value of $f’(a)$?
All I’m asking is how I would begin solving this problem.
 A: In general
$$
f'(a) = \lim_{x\to a} \frac{f(x) - f(a)}{x - a} \tag{1}
$$
With this in mind, rewrite your limit as
$$
\frac{1}{4}\lim_{x\to \pi/4} \frac{\tan x - \tan(\pi/4)}{x - \pi/4} \tag{2}
$$
Can you take it from here?
A: 
HINT, for the limit use:
$$\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}\tag1$$


So, we get:
$$\lim_{x\to\frac{\pi}{4}}\space\frac{\tan\left(x\right)-1}{4x-\pi}=\lim_{x\to\frac{\pi}{4}}\space\frac{\frac{\sin\left(x\right)}{\cos\left(x\right)}-1}{4x-\pi}=\lim_{x\to\frac{\pi}{4}}\space\frac{\cos\left(x\right)-\sin\left(x\right)}{\cos\left(x\right)\cdot\left(\pi-4x\right)}=$$
$$\left\{\lim_{x\to\frac{\pi}{4}}\space\frac{1}{\cos\left(x\right)}\right\}\cdot\left\{\lim_{x\to\frac{\pi}{4}}\space\frac{\cos\left(x\right)-\sin\left(x\right)}{\pi-4x}\right\}=$$
$$\left\{\lim_{x\to\frac{\pi}{4}}\space\frac{1}{\cos\left(x\right)}\right\}\cdot\left\{\lim_{x\to\frac{\pi}{4}}\space\frac{\frac{\text{d}}{\text{d}x}\left(\cos\left(x\right)-\sin\left(x\right)\right)}{\frac{\text{d}}{\text{d}x}\left(\pi-4x\right)}\right\}=$$
$$\left\{\lim_{x\to\frac{\pi}{4}}\space\frac{1}{\cos\left(x\right)}\right\}\cdot\left\{\lim_{x\to\frac{\pi}{4}}\space\frac{\cos\left(x\right)+\sin\left(x\right)}{4}\right\}\tag2$$
A: The way you would BEGIN solving this problem is to bear in mind that
$$
f'(a) = \lim_{x\to a} \frac{f(x)-f(a)}{x-a}.
$$
In the denominator you have $4x-\pi,$ so you can write
$$
\frac{\quad\cdots\cdots\quad}{4x-\pi} \quad = \quad  \frac 1 4 \cdot \frac{\quad\cdots\cdots\quad}{x - \frac \pi 4}.
$$
That should suggest what number $a$ is.
