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

karush

Well-known member
Jan 31, 2012
2,775
$\displaystyle\frac{d}{dx} sec^2(2x)-tan^2(2x)$

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
Jan 26, 2012
1,404
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 [tex]\displaystyle \begin{align*} \sec^2{(x)} \equiv 1 + \tan^2{(x)} \end{align*}[/tex], giving

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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
Aug 30, 2012
1,140
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:
\(\displaystyle \frac{d}{dx} \left ( sec^2(2x) \right ) - tan^2(2x)\)

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

-Dan