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I was wondering whether there was a nice formula for something like

$$ \dfrac{\partial}{\partial x} \left| e^x + (1+i)e^{-x} \right|. $$

(Note that the function is chosen on purpose to have no discontinuities in the derivative, as the argument to the absolute value function never passes through zero.)

So I typed into WolframAlpha:

D[|exp(x) + (1+i)*exp(-x)|, x]

and it said:

Purported solution and plot

Huh!? What imaginary component?

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With Wolfram Alpha, you get what you pay for.

If $f(x)=|e^x+(1+i)e^{-x}|$ then $$F(x)=f(x)^2=(e^x+e^{-x})^2+e^{-2x}=e^{2x}+2+2e^{-2x}.$$ Then $F'(x)=2f(x)f'(x)$ and so $$f'(x)=\frac{F'(x)}{2f(x)}=\frac{e^{2x}-2e^{-2x}}{f(x)}.$$

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Well from what I know, $|e^x+(1+i)e^{-x}|=\sqrt{(e^x+e^{-x})^2+e^{-2x}}=\sqrt{e^{2x}+2+2e^{-2x}}$.

Take the derivative of that to get $\dfrac{\partial}{\partial x}|e^x+(1+i)e^{-x}|=\boxed{\displaystyle \frac{e^{2x}-2e^{-2x}}{\sqrt{e^{2x}+2+2e^{-2x}}}}$.

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  • $\begingroup$ because $|x| = \sqrt{[re(x)]^2+[im(x)]^2}$ $\endgroup$ – Saketh Malyala Jun 9 '17 at 6:12

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