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I am trying to find the derivative of $$\large e^{2 \pi i t \sin(\pi/(2t))}.$$

I know that I am to take the derivative of the exponent, and then multiply it by the beginning problem - the piece that is giving me trouble is the $i$, I think.

Help is appreciated!

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    $i$ is just another constant.2012-12-08
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    Does $\pi/2t$ mean $\frac{\pi}{2t}$ or $\frac{\pi}{2}t$ (in other words, $\pi/(2t)$ or $(\pi/2)t$)?2012-12-08
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    Sorry - pi/(2t).2012-12-08

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By the chain rule, $$\frac{d}{dt}e^{u}=\frac{du}{dt}e^{u}$$ You have $u=2\pi it \sin(\pi/2t)$, so by the product rule $$\frac{du}{dt}=2\pi i \sin(\pi/2t)+2\pi i t \frac{d}{dt}\sin(\pi/2t)$$ Now for this derivative, let $v=\frac{\pi}{2t}$ so $\frac{dv}{dt}=-\frac{\pi}{2t^{2}}$. Use the chain rule to get $$\frac{d}{dt}\sin(v)=\cos(v)\frac{dv}{dt}=\cos(\pi/2t)\cdot \frac{-\pi}{2t^{2}}$$ Substituting back in gives $$\frac{du}{dt}=2\pi i \sin(\pi/2t)-2\pi i t \cos(\pi/2t)\cdot \frac{\pi}{2t^{2}}$$ Therefore $$\frac{d}{dt}e^{2\pi i t \sin(\pi/2t)}=2\pi i \left(\sin(\pi/2t)-\frac{\pi}{2t}\cos(\pi/2t)\right)e^{2\pi i t \sin(\pi/2t)}$$

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    Oh dear. $\pi/2t$ meant $\frac{\pi}{2}t$, didn't it?2012-12-08
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    No, it meant pi/(2t).2012-12-08
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    That's alright then!2012-12-08
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Differentiate complex functions the same way you would with real functions. Treat $i$ as a constant.

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    But why? ${}{}{}{}$2012-12-08
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    That is not the function given in the question.2012-12-08
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    @JonasMeyer Thanks, fixed2012-12-08