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Calculus1 7 Online
OpenStudy (anonymous):

r=sin(4-2theta)

OpenStudy (anonymous):

\[Find dr/d \theta. r=\sin (4-2\theta)\]

OpenStudy (anonymous):

@zepdrix

OpenStudy (anonymous):

ambra te paundi hai udariyan

zepdrix (zepdrix):

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. :)

OpenStudy (anonymous):

phul koi vilayti ala lae gaya udd main reh gaya kyarian

OpenStudy (anonymous):

cos theta

OpenStudy (anonymous):

ur previous question was solved @lovesit2x

OpenStudy (anonymous):

ho attiye kapurthaliye

OpenStudy (anonymous):

yes it was, thanks for your help @mathsmind

zepdrix (zepdrix):

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. \[\huge r'=\cos(4-2\theta)\color{royalblue}{\left(4-2\theta\right)'}\]

OpenStudy (anonymous):

cos(4-2theta) (-2) ?

zepdrix (zepdrix):

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 (zepdrix):

Yah sounds good!

OpenStudy (anonymous):

:D yay!!

zepdrix (zepdrix):

\[\huge r'=-2\cos(4-2\theta)\]Yay!

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