How do you find the derivative of #(sin x)^(e^x)#?

1 Answer
Eddie
Jun 23, 2016

# (sin x)^(e^x) e^x( ln (sin x) + cot x)#

Explanation:

#y =(sin x)^(e^x)#

#ln y =ln ((sin x)^(e^x) )= e^x ln (sin x)#

#1/y y' = e^x ln (sin x) + e^x (ln (sin x))'#

#(ln (sin x))' = 1/(sin x) cos x#

#\implies y' = (sin x)^(e^x) (e^x ln (sin x) + e^x 1/(sin x) cos x)#

#= (sin x)^(e^x) e^x( ln (sin x) + cot x)#

Related questions
See all questions in Chain Rule
Impact of this question
2120 views around the world
You can reuse this answer
Creative Commons License