What is the derivative of #f(theta)=e^(sin2theta)# ?
1 Answer
Aug 3, 2014
#f'(theta)=2cos(2theta)e^sin(2theta)# Explanation,
let's we have
#u(theta)=e^g(f(theta))# then, using Chain Rule ,
#u'(theta)=e^g(f(theta))(g'f(theta))f'(theta)# Similarly differentiating the given function with respect to
#theta# ,
#f(theta)=e^sin(2theta)#
#f'(theta)=e^sin(2theta)(cos(2theta))(2)#
#f'(theta)=2cos(2theta)e^sin(2theta)#
