1. f(theta)= sec (theta) ^2

The answer is 2 theta sec^2 tan ^2 theta.

But I got 2sec ^2 theta tan theta.

2. y= sin^2 x- cos^2 x

The answer is 2sin2x.

But I got 4 sinxcosx.

I don't know how to get to these answers. Can someone please help me?

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- Oct 25th 2009, 04:48 PMPower of OnePleae help me find these derivatives!
**1. f(theta)= sec (theta) ^2**

The answer is 2 theta sec^2 tan ^2 theta.

But I got 2sec ^2 theta tan theta.

2. y= sin^2 x- cos^2 x

The answer is 2sin2x.

But I got 4 sinxcosx.

I don't know how to get to these answers. Can someone please help me? - Oct 25th 2009, 04:55 PMChris L T521
- Oct 25th 2009, 04:58 PMPower of One
- Oct 25th 2009, 05:01 PMpickslides
- Oct 25th 2009, 05:01 PMChris L T521
If that is the case, then by chain rule, we get

$\displaystyle f^{\prime}\!\left(\theta\right)\sec\!\left(\theta^ 2\right)\tan\!\left(\theta^2\right)\cdot2\theta=2\ theta\sec\!\left(\theta^2\right)\tan\!\left(\theta ^2\right)$.

To answer your second question, its a double angle identity: $\displaystyle \sin\!\left(2\theta\right)=2\sin\theta\cos\theta$.

From you answer, it followed that $\displaystyle 4\sin\theta\cos\theta=2\left(2\sin\theta\cos\theta \right)=2\sin\!\left(2\theta\right)$.

Does this clarify things? - Oct 25th 2009, 05:03 PMpickslides
- Oct 25th 2009, 05:09 PMPower of One
**Thank you guys for reminding me to memorize my trig. identities and double-angle formulas!**

@Chris L T521- I still don't understand the first problem. Using the chain rule,

I would bring 2 to the front minus one a power from sec theta ^2 to be left with sec theta, then multiply by its derivative sec theta tan theta.

What am I doing wrong here? :( - Oct 25th 2009, 05:17 PMChris L T521
That's a pitfall you should stay away from: $\displaystyle \sec^2\theta\neq\sec\!\left(\theta^2\right)$

To make it a little easier, let $\displaystyle u=\theta^2$.

So when we differentiate $\displaystyle \sec u$ with respect to theta, we have:

$\displaystyle f^{\prime}\!\left(\theta\right)=\sec u\tan u\cdot\frac{\,du}{\,d\theta}$

Since $\displaystyle u=\theta^2,\,\frac{\,du}{\,d\theta}=2\theta$.

So, $\displaystyle f^{\prime}\!\left(\theta\right)=\sec\!\left(\theta ^2\right)\tan\!\left(\theta^2\right)\cdot 2\theta=2\theta\sec\!\left(\theta^2\right)\tan\!\l eft(\theta^2\right)$.

Does this clarify things? - Oct 25th 2009, 05:25 PMPower of One
- Oct 25th 2009, 05:27 PMChris L T521
- Oct 25th 2009, 05:30 PMPower of One
**I FINALLY UNDERSTAND IT! Thank you so much for being so patient with my stupid self. I really appreciate the help!**