Posted by **carolin** on Sunday, October 18, 2009 at 6:47pm.

how do not understand how to find a deriviative of this function?Do i use chain rule and product rule?

r= the square root of (theta*sin*theta)

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- calculus -
**bobpursley**, Sunday, October 18, 2009 at 6:51pm
ok, if f(u) exists, and u=G(x)

then

y= f(u)

dy/dx= d(f(u)/du * du/dx

but u=G(x)

so du/dx= G'

Now in practice.

r= sqrt(Theta*sinTheta)

dr/dtheta= 1/2 (1/sqrt(Theta*sinTheta))* d(theta*sinTheta)/dtheta

and d(theta*sinTheta)/dtheta= sinTheta+ theta*cosTheta.

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