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Find dy/dx at the given point.
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\[\cos^2(9y)=x+y\] at points \[(\frac{ 2-\pi }{ 4 }) (\frac{ \pi }{ 4 })\]
\[-2\cos(9y)\times \sin(9y)\times 9y'=1+y'\]
you applied the chain rule to cos?
i guess it is better to write \[-18\cos(9y)\sin(9y)y'=1+y'\] yes it is the chain rule repeatedly
derivative of something squared is two times something derivative of cosine is - sine derivative of \(9y\) is \(9y'\) as you are considering \(y\) as a function of \(x\) i.e. \(y=y(x)\)
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so now i factor out my y' from the left hand side?
\[y'=\frac{ 1+y' }{ -18\cos(9y)\sin(9y) }\]
confused because if i take my y' and move it to the left hand side, im left with zero which doesnt make sense.
:(
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