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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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    Sorry - $p$i/(2t).2012-12-08

2 Answers 2

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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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    $T$hat'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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    @JonasMeyer Thanks, fixed2012-12-08