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Mathematics 18 Online
OpenStudy (anonymous):

Find the derivative of r(θ) = θ^2(sinθ) + 2θcosθ

OpenStudy (p0sitr0n):

use the chain rule+product rule i will write x instead of theta r'(x)=2xsinx+x^2cosx+xcosx-2xsinx

OpenStudy (p0sitr0n):

using the fact that h(x)=g(x)j(x) h'(x)=g'(x)j(x)+g(x)j'(x)

OpenStudy (anonymous):

How did you get xcosx? Isn't d/dx 2x=2?

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