MHB Derivative of sec²(2x) - tan²(2x)

[karush]

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

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]

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]

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:

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

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

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

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

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

 

[topsquark]

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

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

Please use parenthesis when needed. The line above reads:
[math]\frac{d}{dx} \left ( sec^2(2x) \right ) - tan^2(2x)[/math]

not
[math]\frac{d}{dx} \left ( sec^2(2x) - tan^2(2x) \right )[/math]
which is what you wanted.

-Dan