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Math Help - Quick question about differentiation of sin

  1. #1
    Senior Member Paze's Avatar
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    Quick question about differentiation of sin

    How come dy/dx \sin^2(x) is \sin(2x)? Thanks.
    Last edited by Paze; April 24th 2013 at 12:45 PM.
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    MHF Contributor MarkFL's Avatar
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    Re: Quick question about differentiation of sin

    What do you get when you differentiate?
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    Senior Member Paze's Avatar
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    Re: Quick question about differentiation of sin

    Well I did like power rule and chain rule uhh...I'm kinda lost.. I got 2*cos(x) I believe..
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    MHF Contributor MarkFL's Avatar
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    Re: Quick question about differentiation of sin

    Yes the power and chain rules are appropriate here, but you lost sin(x) in your result...

    \frac{d}{dx}\left(\sin^2(x) \right)=2\sin^{2-1}(x)\cdot\frac{d}{dx}(\sin(x))=2\sin(x)\cos(x)

    What is the double-angle identity for sine?
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    Re: Quick question about differentiation of sin

    Quote Originally Posted by Paze View Post
    How come dy/dx \sin^2(x) is \sin(2x)? Thanks.
    \displaystyle \begin{align*} y &= \sin^2{(x)} \\ &= \frac{1}{2} - \frac{1}{2}\cos{(2x)} \\ \\ \frac{dy}{dx} &= 0 - 2\cdot \frac{1}{2} \left[ -\sin{(2x)} \right] \\ &= \sin{(2x)} \end{align*}
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    Senior Member Paze's Avatar
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    Re: Quick question about differentiation of sin

    Thanks Mark. I'm not too good at identifying double angle identities. I was unsure of my method due to the fact that I have a function of x squared. So I'm thinking: ''Is that the power rule or the chain rule?'' Never occured to me that when a function is squared, the function is chain rule and the 'squared' part is power rule. It was intuitive to me, but I didn't quite SEE it. Thanks for clarifying.

    Prove It: although your guidance is usually of SUPERB quality, your description confuses me even more this time. You must be applying some identities in your calculations that I am not aware of (or I could be obtuse, sorry if I am).
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    Re: Quick question about differentiation of sin

    Quote Originally Posted by Paze View Post
    Thanks Mark. I'm not too good at identifying double angle identities. I was unsure of my method due to the fact that I have a function of x squared. So I'm thinking: ''Is that the power rule or the chain rule?'' Never occured to me that when a function is squared, the function is chain rule and the 'squared' part is power rule. It was intuitive to me, but I didn't quite SEE it. Thanks for clarifying.

    Prove It: although your guidance is usually of SUPERB quality, your description confuses me even more this time. You must be applying some identities in your calculations that I am not aware of (or I could be obtuse, sorry if I am).
    My guidance is as superb as it always is, and you are so obtuse you are almost reflex :P but I accept your apology.

    Surely you know the double angle identity for cosine...

    \displaystyle \begin{align*} \cos{(2x)} &= \cos^2{(x)} - \sin^2{(x)} \\ \cos{(2x)} &= 1 - \sin^2{(x)} - \sin^2{(x)} \\ \cos{(2x)} &= 1 - 2\sin^2{(x)} \\ 2\sin^2{(x)} &= 1 - \cos{(2x)} \\ \sin^2{(x)} &= \frac{1}{2} - \frac{1}{2}\cos{(2x)}  \end{align*}

    Btw no offence was intended, it's just some light-hearted ribbing
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    Senior Member Paze's Avatar
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    Re: Quick question about differentiation of sin

    Quote Originally Posted by Prove It View Post
    My guidance is as superb as it always is, and you are so obtuse you are almost reflex :P but I accept your apology.

    Surely you know the double angle identity for cosine...

    \displaystyle \begin{align*} \cos{(2x)} &= \cos^2{(x)} - \sin^2{(x)} \\ \cos{(2x)} &= 1 - \sin^2{(x)} - \sin^2{(x)} \\ \cos{(2x)} &= 1 - 2\sin^2{(x)} \\ 2\sin^2{(x)} &= 1 - \cos{(2x)} \\ \sin^2{(x)} &= \frac{1}{2} - \frac{1}{2}\cos{(2x)}  \end{align*}

    Btw no offence was intended, it's just some light-hearted ribbing
    lol. Thanks for clarifying. I see your point now. The thing is, I just completed trigonometry in school now (final test today. It went very well ) so it's not exactly second nature to me. These identities still have to seep in with time. I think I'm better at calculus than trigonometry, since 90% of my math knowledge is self-taught. But that, of course, means that I have a lot of gaps in my mathematical background!
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    MHF Contributor MarkFL's Avatar
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    Re: Quick question about differentiation of sin

    As you progress in your studies, you will get many opportunities to fortify your knowledge of pre-calculus concepts.
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