Quick question about the derivative of a complex-valued function

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The discussion centers on the differentiation of complex-valued functions, specifically examining the relationship between the real part of a function and its derivative. The user inquires whether the equation \(\frac{d}{dx} Re(f) = Re\left(\frac{df}{dx}\right)\) holds true for a function defined as \(f(x) = u(x) + iv(x)\), where \(u\) and \(v\) are real-valued functions. The consensus confirms that this equation is valid when \(u\) and \(v\) are real functions, simplifying the process of handling derivatives in trigonometric identities.

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tjackson3
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This is something that has been bothering me for awhile. Suppose we have some function f(x) = u(x) + iv(x), where u,v are real-valued functions. Obviously Re(f) = u(x) and Im(f) = v(x), both of which are real. My question is: in general, is it safe to say that

\frac{d}{dx} Re(f) = Re\left(\frac{df}{dx}\right)

? If so, that makes life a lot easier with trig identities dealing with derivatives. This seems really trivial, and I feel like it's correct, but I can't figure out how one would prove it, and I don't want to use it just assuming that it is true...

Thanks!
 
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Hi tjackson3! :smile:

no problem … df(x)/dx = du(x)/dx + idv(x)/dx,

and if u,v (and x) are real-valued functions, then so are du(x)/dx, dv(x)/dx. :wink:
 
tjackson3, is x a real variable?
 

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