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)#