What is derivative of y = 1/(4e^(2x-1)+sin(e^(2x+1)))?

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You need to differentiate the function using the quotient rule and the chain rule, such that:

`y' = (1'*(4e^(2x-1)+sin(e^(2x+1))) - 1*(4e^(2x-1)+sin(e^(2x+1)))')/((4e^(2x-1)+sin(e^(2x+1)))^2)`

`y' = (4*(e^(2x-1))' + (sin(e^(2x+1)))')/((4e^(2x-1)+sin(e^(2x+1)))^2)`

`y' = (4*2(e^(2x-1)) + 2e^(2x-1)*cos(e^(2x+1)))/((4e^(2x-1)+sin(e^(2x+1)))^2)`

`y' = (8(e^(2x-1)) + 2e^(2x-1)*cos(e^(2x+1)))/((4e^(2x-1)+sin(e^(2x+1)))^2)`

Hence, evaluating the derivative of the given composed function, yields `y' = (8(e^(2x-1)) + 2e^(2x-1)*cos(e^(2x+1)))/((4e^(2x-1)+sin(e^(2x+1)))^2).`

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