Differentiating $y=\tan(x)$ Differentiate $$y=\tan(x)$$
normally we write $$y=\frac{\sin(x)}{\cos(x)}$$ and using quotient rule of differentiation
$$\frac{dy}{dx}=\frac{\cos(x)\sin(x)'-\sin(x)\cos(x)'}{\cos^2(x)}$$
$$\frac{dy}{dx}=\frac{\cos^{2}(x)+\sin^{2}(x)}{\cos^{2}(x)}=\sec^2(x)$$.
Question: Is there another alternative way of differentiating $y=\tan(x)?$
 A: Using the tangent addition formula,
$$\tan(\alpha+\beta) = \frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$$
plus the fact that
$$\lim_{h\to 0}\frac{\tan(h)}{h} = 1$$
(which can be derived from the fact that the limit of $\frac{\sin(h)}{h}$ as $h\to 0$ is $1$), we have:
$$\begin{align*}
\frac{d}{dx}\tan(x) &= \lim_{h\to 0}\frac{\tan(x+h) - \tan(x)}{h} = \lim_{h\to 0}\frac{\quad\frac{\tan(x)+\tan(h)}{1-\tan(x)\tan(h)} - \tan(x)}{h}\\
&= \lim_{h\to 0}\frac{\tan(x)+\tan(h) - \tan(x)(1-\tan(x)\tan(h))}{h(1-\tan(x)\tan(h))} \\
&= \lim_{h\to 0}\frac{\tan(x) + \tan(h) - \tan(x) + \tan(h)\tan^2(x)}{h(1-\tan(x)\tan(h))}\\
&= \lim_{h\to 0}\frac{\tan(h) + \tan(h)\tan^2(x)}{h(1-\tan(x)\tan(h))}\\
&= \lim_{h\to 0}\frac{\tan(h) (1 + \tan^2(x))}{h(1-\tan(x)\tan(h))}\\
&= \lim_{h\to 0}\left(\frac{\tan(h)}{h}\right)\left(\frac{1+\tan^2(x)}{1-\tan(x)\tan(h)}\right)\\
&= (1)\left(\frac{1+\tan^2(x)}{1}\right) = 1+\tan^2(x) = \sec^2(x).
\end{align*}
$$
A: Construct the line x = 1, and lines from the origin that make angles $t$ and $t+h$ with the x axis.  They form right triangle with sides $1, \tan t, \sec t$ and $1,\tan (t+h), \sec (t+h)$  and areas $\frac 12 \tan t$, and $\frac 12 \tan (t+h)$
And the difference between these two triangle is a triangle of area $\frac 12 ( \tan (t+h)  - \tan t)$

Note that this area is larger than the area of the section of the circle with radius $\sec t$ and angle measure $h.$ and smaller than the area of the section of the circle with radius $\sec (t+h)$ and angle measure $h.$ 
$\frac 12 h\sec^2 t\le \frac 12 (\tan (t+h) - \tan t) \le \frac 12 h\sec^2 (t+h)$
or 
$\sec^2 t\le \frac {\tan (t+h) - \tan t}{h} \le \frac 12 h\sec^2 (t+h)$
And the expression in the middle looks a lot like the definition of the derivative of $\tan t$
Taking the limit as $h$ goes to 0, our derivative gets squeezed.
A: $\tan(x)=\sin(x) \cdot (\cos(x))^{-1}$. 
Then use the product rule
