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Find the derivative of r(θ) = θ^2(sinθ) + 2θcosθ
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use the chain rule+product rule i will write x instead of theta r'(x)=2xsinx+x^2cosx+xcosx-2xsinx
using the fact that h(x)=g(x)j(x) h'(x)=g'(x)j(x)+g(x)j'(x)
How did you get xcosx? Isn't d/dx 2x=2?
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