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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.

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(Continuity check): On chat you said you thought it was $2\tan x \sec^2 x$. That's exactly right. But that's not quite everything. –  mixedmath May 5 '12 at 1:19
    
@Whoops! I misread the tex part - I'm sorry. –  mixedmath May 5 '12 at 1:21
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2 Answers

up vote 11 down vote accepted

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.

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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 $$ (\sec^2 x)' = [ (\sec x)^2]' =2(\sec x)^1 \cdot \color{maroon}{(\sec x)'}=2\sec x\cdot\color{maroon}{ \sec x\tan x }=2\tan x\, \sec^2 x. $$

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