Question

What is the derivative of #y= (e^(ix)-e^(-ix) )/2#?

Answer 1

`dy/dx = i( (e^(ix)-e^(-ix))/2)`

Explanation:

Use the chain rule:

`dy/dx = (e^(ix)(i) - e^(-ix)(-i))/2`

`= (ie^(ix) + ie^(-ix))/2`

Alternatively

Point out that `y= isinx`, so `dy/dx = icosx`