Tangent Triple Threat

Problem

Consider the function $y=\tan x$. Show that $y''' = 2(1+y^2)(1+3y^2)$.

Figure 1. Graph of the tangent function

Solution

Take the three derivatives:

$y=\tan x$

$y' = \sec^2 x$

$y'' = 2\sec^2 x \tan x$

$y''' = 2\sec^4 x + 4\sec^2 \tan^2 x$.

Substitute $y$ and its third derivative into

$y''' = 2(1+y^2)(1+3y^2)$.

$2\sec^4 x + 4\sec^2 \tan^2 x = 2(1 + \tan^2 x)(1 + 3\tan^2 x)$

$2\sec^2 x (\sec^2 x + 2 \tan^2 x) = 2(1 + \tan^2 x)((1 + \tan^2 x) + 2\tan^2 x)$

Since $1+\tan^2x=\sec^2x$

$2\sec^2 x (\sec^2 x + 2 \tan^2 x) = 2\sec^4 x + 4\sec^2 x\tan^2 x$.

Both sides match indeed!