Trig derivatives I am working through the review sections of Stewart's Calc 7 book and I am at this question
$$\frac{d}{dx} (\tan^2 x) = \frac{d}{dx} (\sec^2 x)$$
It is true or false, I said false but the book says true and I do not understand why.
 A: You could just calculate. 
Using the chain rule:
$$
(\tan^2 x)'= [ (\tan x)^2]' =2 (\tan x)^1 \cdot\color{maroon}{ (\tan x)'}=2\tan x\, \color{maroon}{\sec^2 x  },
$$
and
$$
(\sec^2 x)' = [ (\sec x)^2]'  =2(\sec x)^1 \cdot
\color{maroon}{(\sec x)'}=2\sec x\cdot\color{maroon}{ \sec x\tan x }=2\tan x\, \sec^2 x.
$$
A: It is simply a matter of trignometric identities.
I hope you know that
$$\sin ^2 x + \cos ^2 x = 1$$
Dividing through $\cos ^2 x$ gives
$$\sin ^2 x + \cos ^2 x = 1$$
$$\eqalign{
  & \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \frac{{{{\cos }^2}x}}{{{{\cos }^2}x}} = \frac{1}{{{{\cos }^2}x}}  \cr 
  & {\tan ^2}x + 1 = {\sec ^2}x \cr} $$
This means that
$${\sec ^2}x - {\tan ^2}x = 1$$
Since these differ by a constant, $1$, their derivatives are equal:
$$\eqalign{
  & \frac{d}{{dx}}{\sec ^2}x - \frac{d}{{dx}}{\tan ^2}x = \frac{d}{{dx}}1  \cr 
  & \frac{d}{{dx}}{\sec ^2}x - \frac{d}{{dx}}{\tan ^2}x = 0  \cr 
  & \frac{d}{{dx}}{\sec ^2}x = \frac{d}{{dx}}{\tan ^2}x \cr} $$
Note it is not necessary to calculate the derivatives explicitly, although it might prove useful to do so.
