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?
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?
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?
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?
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?