lovesit2x
r=sin(4-2theta)
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lovesit2x
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\[Find dr/d \theta. r=\sin (4-2\theta)\]
lovesit2x
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@zepdrix
ravina
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ambra te paundi hai udariyan
zepdrix
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Hmm so what would the derivative of \(\large \sin\theta\) be?
This is another one of those problems where we'll have to work with the chain rule. :)
ravina
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phul koi vilayti ala lae gaya udd main reh gaya kyarian
lovesit2x
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cos theta
mathsmind
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ur previous question was solved @lovesit2x
ravina
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ho attiye kapurthaliye
lovesit2x
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yes it was, thanks for your help @mathsmind
zepdrix
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Yes good!
Since the inside of the sine is more than just \(\large \theta\), we have to apply the chain rule, multiplying by the derivative of the inside.
\[\huge r'=\cos(4-2\theta)\color{royalblue}{\left(4-2\theta\right)'}\]
lovesit2x
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cos(4-2theta) (-2) ?
zepdrix
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Understand the process? We took the derivative of sine, which gave us cosine (with the same contents inside of it). And then we make a copy of the inside, and multiply by it's derivative.
zepdrix
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Yah sounds good!
lovesit2x
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:D yay!!
zepdrix
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\[\huge r'=-2\cos(4-2\theta)\]Yay!