How do you calculate the derivative of #r= 2thetasqrt(sec theta)#?

1 Answer
Truong-Son N.
Jul 9, 2015

Here we have #r(theta) = 2thetasqrtsectheta#

#((dr)/(d theta))[f(theta)g(theta)] = f(theta)g'(theta) + g(theta)f'(theta)#

#= (2theta)(1/(2sqrtsectheta)*secthetatantheta) + (sqrtsectheta)(2)#

#= thetasqrtsecthetatantheta + 2sqrtsectheta#

#= 2sqrtsectheta + thetasqrtsecthetasintheta/costheta#

#= color(blue)(2sqrtsectheta + thetasinthetasec^"3/2"theta)#

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