Question

What is the derivative of `f(x) = cos(ix)+sin(ix)`?

Answer 1

`f'(x)=i(cos(ix)-sin(ix)) = sinh(x)+icosh(x)`

Explanation:

We will use the following:

• The sum rule

• The chain rule

• `d/dx cos(x) = -sin(x)`

• `d/dx sin(x) = cos(x)`

• `d/dx ax = a` for `a in CC`

`f'(x) = d/dx(cos(ix)+sin(ix))`

`=d/dxcos(ix) + d/dxsin(ix)`

`=-sin(ix)(d/dxix) + cos(ix)(d/dxix)`

`=-sin(ix)(i) + cos(ix)(i)`

`=i(cos(ix)-sin(ix))`

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We could also go about this by using hyperbolic trig functions and the following:

• `-isin(ix) = sinh(x)`
• `cos(ix) = cosh(x)`
• `d/dx sinh(x) = cosh(x)`
• `d/dx cosh(x) = sinh(x)`

Then:

`d/dx(cos(ix)+sin(ix)) = d/dx(cosh(x) + isinh(x))`

`=d/dxcosh(x)+d/dxisinh(x)`

`=sinh(x)+icosh(x)`

Note that substituting the standard trig functions back in gives the same result as obtained above.