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I would like help with this problem.

Prove this is true by using complex analysis. This problem appeared in my workbook and there appears to be no solution in the back.

$$\frac{d}{dx}e^{tx} = te^{tx}.$$

Edit: I think I am supposed to use chain rule?

Quaxton Hale
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  • So $x$ and $t$ are complex numbers and $d/dx$ denotes complex derivative? Yes, this follows for instance from the chain rule. – Julien Feb 28 '13 at 03:46
  • Okay, thank you. I wasn't sure if using the chain rule was complex analysis or just calculus. Edit: It says I need to use a different formula or method to prove this true. – Quaxton Hale Feb 28 '13 at 03:53
  • Possibly Euler's formula? – Quaxton Hale Feb 28 '13 at 04:04
  • I feel uncomfortable trying to intuit the desires of whoever is asking you this question, but perhaps they want you to use something like Cauchy's integral formula? – Antonio Vargas Feb 28 '13 at 04:10
  • @AntonioVargas Presumably they wanted a weaker result to be used, not a stronger one. – EuYu Feb 28 '13 at 04:11
  • @Justin You should probably mention what you've learned so far. The chain rule is by far the most straightforward way to this. If this is an exercise in a book, then perhaps tell us what was covered in the chapter where the question was posed. We're not going to be able to guess how you want the problem solved. – EuYu Feb 28 '13 at 04:15
  • Hi @EuYu, we've started proving trigonometric functions with Euler's formula: ei t = cos t + i sin t I think this is how I should prove this to be true. By taking the derivative of cos(t) + isin(t) ? – Quaxton Hale Feb 28 '13 at 04:24
  • @Justin Presumably you've only covered Euler's formula for real numbers? – EuYu Feb 28 '13 at 04:28
  • @Justin The question's title is not very helpful. Perhaps you could change it? Maybe something like "Derivative of the complex exponential" – Tyler Feb 28 '13 at 04:31
  • Your answer will also depend on how you've defined the exponential function. If $e^x = e^{\operatorname{re}(x)}e^{i\operatorname{im}(x)}$, the answer will be different than if $e^x = \sum \frac{x^n}{n!}$. – Neal Feb 28 '13 at 04:46

1 Answers1

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If you don't mind, I will use the complex number letters I am used to.

So a complex function of the complex variable is holomorphic at $z_0$ if $$ \lim_{z\rightarrow z_0}\frac{f(z)-f(z_0)}{z-z_0} $$ exists in $\mathbb{C}$. In this case, it is denoted by $f'(z_0)$.

This notion obeys the usual rules you are used to for derivatives of real functions of the real variable. In particular, there is a chain rule. But note there is much much more structure on holomorphic functions, than on differentiable real functions of the real variable.

So the function you are considering is $$ g(z):=e^{uz}. $$ This is the composition $g=\exp\circ f$ with $f(z)=uz$. Now $\exp$ is holomorphic on $\mathbb{C}$ with derivative equal to itself (this is a fact which follows from the definition of $\exp$ as the sum of the series $\sum_{n\geq 0} z^n/n!$), and $f$ is holomorphic on $\mathbb{C}$ with derivative equal to $u$ (this is easy from the definition).

Now by the chain rule, $g$ is holomorphic on $\mathbb{C}$ and $$ g'(z)=\exp'(f(z))f'(z)=e^{uz}u. $$

Julien
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