Pleae help me find these derivatives!

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October 25th 2009, 04:48 PM
Power of One

Pleae 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?
October 25th 2009, 04:55 PM
Chris L T521

Quote:

Originally Posted by Power of One

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?

For the first one, is it $\sec\!\left(\theta^2\right)$ or $\sec^2\!\left(\theta\right)$ ?

Your second answer is correct, since $4\sin x\cos x=2(2\sin x\cos x)=2\sin\!\left(2x\right)$
October 25th 2009, 04:58 PM
Power of One

Quote:

Originally Posted by Chris L T521

For the first one, is it $\sec\!\left(\theta^2\right)$ or $\sec^2\!\left(\theta\right)$ ?

Your second answer is correct, since $4\sin x\cos x=2(2\sin x\cos x)=2\sin\!\left(2x\right)$

The first one is $\sec\!\left(\theta^2\right)$ .

Can you explain to me how you got $4\sin x\cos x=2(2\sin x\cos x)=2\sin\!\left(2x\right)$ ?
October 25th 2009, 05:01 PM
pickslides

Quote:

Originally Posted by Power of One

But I got 4 sinxcosx.

$4\sin(x)\cos(x) = 2\times 2 \sin(x)\cos(x) = 2\times \sin(2x)$
October 25th 2009, 05:01 PM
Chris L T521

Quote:

Originally Posted by Power of One

The first one is $\sec\!\left(\theta^2\right)$ .

Can you explain to me how you got $4\sin x\cos x=2(2\sin x\cos x)=2\sin\!\left(2x\right)$ ?

If that is the case, then by chain rule, we get

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

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

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

Does this clarify things?
October 25th 2009, 05:03 PM
pickslides

Quote:

Originally Posted by Power of One

Can you explain to me how you got $4\sin x\cos x=2(2\sin x\cos x)=2\sin\!\left(2x\right)$ ?

It is a known trig identity. You should commit it to memory.
October 25th 2009, 05:09 PM
Power 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? :(
October 25th 2009, 05:17 PM
Chris L T521

Quote:

Originally Posted by Power 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? :(

That's a pitfall you should stay away from: $\sec^2\theta\neq\sec\!\left(\theta^2\right)$

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

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

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

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

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

Does this clarify things?
October 25th 2009, 05:25 PM
Power of One

Quote:

Originally Posted by Chris L T521

That's a pitfall you should stay away from: $\sec^2\theta\neq\sec\!\left(\theta^2\right)$

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

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

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

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

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

Does this clarify things?

I'm sorry, but no it doesn't. I tried to follow you, but I got lost. Where did du/ d theta come from?
October 25th 2009, 05:27 PM
Chris L T521

Quote:

Originally Posted by Power of One

I'm sorry, but no it doesn't. I tried to follow you, but I got lost. Where did du/ d theta come from?

Its a result from the chain rule:

$\frac{\,df}{\,d\theta}=\frac{\,df}{\,du}\frac{\,du }{\,d\theta}$

where, in our problem, $f=\sec u$ and $u=\theta^2$
October 25th 2009, 05:30 PM
Power of One

I FINALLY UNDERSTAND IT! Thank you so much for being so patient with my stupid self. I really appreciate the help!