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Math Help - Pleae help me find these derivatives!

  1. #1
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    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?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Power of One View Post
    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)
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    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)?
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  4. #4
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    Quote Originally Posted by Power of One View Post

    But I got 4 sinxcosx.
    4\sin(x)\cos(x) = 2\times 2 \sin(x)\cos(x) = 2\times \sin(2x)
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Power of One View Post
    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?
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  6. #6
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    Quote Originally Posted by Power of One View Post

    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.
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  7. #7
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    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?
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  8. #8
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Power of One View Post
    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?
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  9. #9
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    Quote Originally Posted by Chris L T521 View Post
    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?
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  10. #10
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Power of One View Post
    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
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  11. #11
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    I FINALLY UNDERSTAND IT! Thank you so much for being so patient with my stupid self. I really appreciate the help!
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