$$\dfrac{dy}{dx} =\dfrac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\dfrac{f(\theta) \cos \theta +f'(\theta) \sin \theta}{-f(\theta) \sin \theta +f'(\theta) \cos \theta} \\ r=f(\theta), ~~y=f(\theta)\sin \theta,~~x=f(\theta)\cos \theta\\r=\theta$$ This is the formula for the first derivative of polar equations. Using the second line making a few substitutions to find the first derivative we find that $$\dfrac{dy}{d\theta} =\dfrac{\theta \cos \theta +\sin \theta}{-\theta\sin \theta +\cos \theta}$$ which according to te model answer is correct. It is from my understand that the second derivative is found as $$\dfrac{d}{d\theta} \cdot \dfrac{dy}{d\theta}=\dfrac{dy}{d\theta\cdot d\theta}=\dfrac{\theta \cos \theta +\sin \theta}{(-\theta\sin \theta +\cos \theta)^2}$$ however i'm now sure how i would arrive at the model answer of $$\dfrac{\theta ^2 +2}{(-\theta\sin \theta +\cos \theta)^3}$$ Can anyone guide me along the way on how to work this out? Thanks.
$\endgroup$ 12 Reset to defaultFinding the second derivative of a polar equation
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