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