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What is the derivative of #f(theta)=e^(sin2theta)# ?

Answer 1

Emma Aldridge

#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)#

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Answer 2

Christopher Agresta

The derivative of ( f(\theta) = e^{\sin(2\theta)} ) is ( f'(\theta) = 2e^{\sin(2\theta)} \cos(2\theta) ).

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Answer from HIX Tutor

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