Differentiate $$y=\tan(x)$$

normally we write $$y=\frac{\sin(x)}{\cos(x)}$$ and using quotient rule of differentiation



Question: Is there another alternative way of differentiating $y=\tan(x)?$

  • $\begingroup$ Writing it as $\sec ^2(x) -1$ and differentiating would work. $\endgroup$ – JacobCheverie Apr 30 at 21:42
  • $\begingroup$ Maybe using another trig identity expressing $\tan x$ such as $\tan x=[\sin(2x)]/[1+\cos(2x)]$ but it would be longer. $\endgroup$ – coffeemath Apr 30 at 21:45
  • $\begingroup$ Do you mean another way to establish the formula, or another (equivalent) final formula? $\endgroup$ – Bernard Apr 30 at 21:54
  • $\begingroup$ What sort of "other way" do you mean? $\endgroup$ – The Count Apr 30 at 21:54
  • $\begingroup$ like using infinite series or ... $\endgroup$ – user550260 Apr 30 at 21:56

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: $$\begin{align*} \frac{d}{dx}\tan(x) &= \lim_{h\to 0}\frac{\tan(x+h) - \tan(x)}{h} = \lim_{h\to 0}\frac{\quad\frac{\tan(x)+\tan(h)}{1-\tan(x)\tan(h)} - \tan(x)}{h}\\ &= \lim_{h\to 0}\frac{\tan(x)+\tan(h) - \tan(x)(1-\tan(x)\tan(h))}{h(1-\tan(x)\tan(h))} \\ &= \lim_{h\to 0}\frac{\tan(x) + \tan(h) - \tan(x) + \tan(h)\tan^2(x)}{h(1-\tan(x)\tan(h))}\\ &= \lim_{h\to 0}\frac{\tan(h) + \tan(h)\tan^2(x)}{h(1-\tan(x)\tan(h))}\\ &= \lim_{h\to 0}\frac{\tan(h) (1 + \tan^2(x))}{h(1-\tan(x)\tan(h))}\\ &= \lim_{h\to 0}\left(\frac{\tan(h)}{h}\right)\left(\frac{1+\tan^2(x)}{1-\tan(x)\tan(h)}\right)\\ &= (1)\left(\frac{1+\tan^2(x)}{1}\right) = 1+\tan^2(x) = \sec^2(x). \end{align*} $$

  • $\begingroup$ Dang, beat me to it. :-) $\endgroup$ – Brian Tung Apr 30 at 21:58

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 $\frac 12 ( \tan (t+h) - \tan t)$

enter image description here

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 $h.$

$\frac 12 h\sec^2 t\le \frac 12 (\tan (t+h) - \tan t) \le \frac 12 h\sec^2 (t+h)$


$\sec^2 t\le \frac {\tan (t+h) - \tan t}{h} \le \frac 12 h\sec^2 (t+h)$

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.


$\tan(x)=\sin(x) \cdot (\cos(x))^{-1}$. Then use the product rule

  • $\begingroup$ That's the same as doing the quotient rule... $\endgroup$ – Arturo Magidin Apr 30 at 21:50
  • $\begingroup$ He wanted a different way of differentiating, and some people find the product rule easier even though yes, the two are both sides of the same coin. $\endgroup$ – ItIsLastThursday Apr 30 at 21:50

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