Derivative Riddle 2

Calculus Level 2

Img = imaginary part

I m g ( d d x ( − x ) x ∣ 1 ) = ? Img(\frac{d}{dx}(-x)^{x}|_{1})=?

π \pi π + 1 \pi+1 -1 1 0 − π -\pi π − 1 \pi-1

Bar Hemo by Bar Hemo , 26, Israel

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1 solution

Bar Hemo
Mar 17, 2016

( − x ) x (-x)^{x} = ( e − j π x ) x (e^{-j\pi }x)^x = ( e − j π e l n x ) x (e^{-j\pi }e^{ln x})^x = e − j π x + x l n x e^{-j\pi x+xlnx}

d d x ( − x ) x ∣ 1 \frac{d}{dx}(-x)^{x}|_{1} = d d x e − j π x + x l n x ∣ 1 \frac{d}{dx}e^{-j\pi x+xlnx}|_{1} = ( − j π + l n x + 1 ) e − j π x + x l n x ∣ 1 (-j\pi+lnx+1)e^{-j\pi x+xlnx}|_{1} =

= ( − j π + 1 ) ( − 1 ) (-j\pi+1)(-1) = j π − 1 j\pi-1

so

I m g ( d d x ( − x ) x ∣ 1 ) Img(\frac{d}{dx}(-x)^{x}|_{1}) = I m g ( j π − 1 ) Img(j\pi-1) = π \pi

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( a x ) y ≠ a ( x y ) i f a ∉ R (a^x)^y\neq a^{(xy)}\quad if \quad a\notin\mathbb R

A Former Brilliant Member - 5 years, 2 months ago

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can you prove it ?

Bar Hemo - 5 years, 2 months ago

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Consider i i i^i .

Note that i = e i π 2 = e i 5 π 2 ⟹ i i = e − π 2 = e − 5 π 2 \large i=e^{i\frac{\pi}{2}}=e^{i\frac{5\pi}{2}}\\\implies \large i^i=e^{\frac{-\pi}{2}}=e^{\frac{-5\pi}{2}}

The last equality has the following implication: e 2 π = 0 e^{2\pi}=0 which is trivially false. This counterexample proves my claim.

For further details, refer Failure of power and logarithm identities

A Former Brilliant Member - 5 years, 2 months ago

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