# [SOLVED]derivative of sec²(2x) - tan²(2x)

#### karush

##### Well-known member
$\displaystyle\frac{d}{dx} sec^2(2x)-tan^2(2x)$

rewriting...

$\displaystyle\frac{1}{cos^2(2x)}-\frac{sin^2(2x)}{cos^2(2x)}$

$\displaystyle\frac{cos^2(2x)}{cos^2(2x)}=1$

$\displaystyle\frac{d}{dx}(1)=0$

I put this in $W|A$ but there were many complicated steps just to get the same answer

how do you use the chain rule on $sec^2(2x)$ and $tan^2(2x)$

#### Prove It

##### Well-known member
MHB Math Helper
Re: derivative of sec^2(2x)-tan^2(2x)

$\displaystyle\frac{d}{dx} sec^2(2x)-tan^2(2x)$

rewriting...

$\displaystyle\frac{1}{cos^2(2x)}-\frac{sin^2(2x)}{cos^2(2x)}$

$\displaystyle\frac{cos^2(2x)}{cos^2(2x)}=1$

$\displaystyle\frac{d}{dx}(1)=0$

I put this in $W|A$ but there were many complicated steps just to get the same answer

how do you use the chain rule on $sec^2(2x)$ and $tan^2(2x)$
What you've done is fine, I don't know why you're worrying.

An alternative is to use the identity \displaystyle \begin{align*} \sec^2{(x)} \equiv 1 + \tan^2{(x)} \end{align*}, giving

\displaystyle \begin{align*} \sec^2{(2x)} - \tan^2{(2x)} &\equiv 1 + \tan^2{(2x)} - \tan^2{(2x)} \\ &\equiv 1 \end{align*}

#### MarkFL

Staff member
Re: derivative of sec^2(2x)-tan^2(2x)

...
how do you use the chain rule on $sec^2(2x)$ and $tan^2(2x)$
Suppose you do not smplify the expression using trigonometric identities before differentiating:

$$\displaystyle \frac{d}{dx}\left(\sec^2(2x)-\tan^2(2x) \right)=$$

Differentiating term by term, using the power and chain rules, we find:

$$\displaystyle 2\sec(2x)\cdot\frac{d}{dx}(\sec(2x))-2\tan(2x)\cdot\frac{d}{dx}(\tan(2x))=$$

$$\displaystyle 2\sec(2x)\cdot\sec(2x)\tan(2x)\cdot\frac{d}{dx}(2x)-2\tan(2x)\cdot\sec^2(2x)\cdot\frac{d}{dx}(2x)=$$

$$\displaystyle 4\sec^2(2x)\tan(2x)-4\sec^2(2x)\tan(2x)=0$$

#### topsquark

##### Well-known member
MHB Math Helper
Re: derivative of sec^2(2x)-tan^2(2x)

$\displaystyle\frac{d}{dx} sec^2(2x)-tan^2(2x)$
$$\displaystyle \frac{d}{dx} \left ( sec^2(2x) \right ) - tan^2(2x)$$
$$\displaystyle \frac{d}{dx} \left ( sec^2(2x) - tan^2(2x) \right )$$