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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