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\[Find dr/d \theta. r=\sin (4-2\theta)\]
ambra te paundi hai udariyan

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Other answers:

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. :)
phul koi vilayti ala lae gaya udd main reh gaya kyarian
cos theta
ur previous question was solved @lovesit2x
ho attiye kapurthaliye
yes it was, thanks for your help @mathsmind
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. \[\huge r'=\cos(4-2\theta)\color{royalblue}{\left(4-2\theta\right)'}\]
cos(4-2theta) (-2) ?
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.
Yah sounds good!
:D yay!!
\[\huge r'=-2\cos(4-2\theta)\]Yay!

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