What is $\frac{d}{dx}x^i$?

Question

What is $\frac{d}{dx}x^i$ where $i=\sqrt{-1}$? Does the proof as for real indices apply in the same way here?

The question needs some clarification. 1) What is your definition of $x^i$? 2) Is $x$ considered to be real (positive)? — mickep, Jan 01 '20 at 17:28

Riz222 · Accepted Answer · 2020-01-01T17:59:35.533

6

The rule does apply, so the derivative is $ix^{i-1}$.

For a simple proof,

$\dfrac{d}{dx}x^i$

$=\dfrac{d}{dx}e^{i\ln x}$

$=e^{i\ln x}\dfrac{d}{dx}i\ln x$

$= e^{i\ln x}\cdot\dfrac{i}{x}$

$ =i\dfrac{e^{i\ln x}}{e^{\ln x}} $

$=ie^{(i-1)\ln x}=ix^{i-1}$

edited Jan 01 '20 at 17:59

answered Jan 01 '20 at 16:46

Riz222

4

Which proof are you thinking of which goes through for complex exponents? – Kevin Carlson Jan 01 '20 at 16:56
Right, so the same proof does not go through. This is not the usual proof for real exponents. – Kevin Carlson Jan 01 '20 at 17:30
This should probably include some statement on the choice of branch. – cmk Jan 01 '20 at 17:34
1

@Kevin Carlson, I admit I didn't initially interpret "does the proof apply" (usually it's rules that apply) as "is the proof valid"... edited now. – Riz222 Jan 01 '20 at 18:01

score 1 · Answer 2 · answered Jan 01 '20 at 16:47

1

$i$ is just treated as a constant, so the power rule applies.

answered Jan 01 '20 at 16:47

2 Answers2