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 ... $\endgroup$ – coffeee 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.