Derivatives with Surprisingly Imaginary Results
Here are three derivatives of functions where complex numbers are involved with further algebraic simplification.
d/dx √(a - x)^(1/2)
d/dx √(a- x)^(1/2)
= 1/2 ∙ (a - x)^(-1/2) ∙ -1
= -1/2 ∙ 1 ÷ (√(a - x))
Going a step further...
= -1/2 ∙ 1 ÷ (√(-1) ∙ √(x - a))
With √(-1) = i , 1/i = -i
= i ÷ (2 ∙ √(x - a))
d/dx arcsin(x + a)
d/dx arcsin(x + a)
= 1 ÷ √(1 - (x + a)^2)
= 1 ÷ √(1 - (x^2 + 2 ∙ a ∙ x + a^2))
= 1 ÷ √(-x^2 - 2 ∙ a ∙ x + 1 - a^2)
Factoring out -1 in the denominator:
= 1 ÷ √((-1) ∙ (x^2 + 2 ∙ a ∙ x - 1 + a^2))
= 1 ÷ (i ∙ √(x^2 + 2 ∙ a ∙ x - 1 + a^2))
= -i ÷ √(x^2 + 2 ∙ a ∙ x - 1 + a^2)
d/dx e^(√(a - x))
d/dx e^(√(a - x))
= e^(√(a - x)) ∙ d/dx √(a - x)
= -e^(√(a - x)) ÷ (2 ∙ √(a - x))
With: √(a - x) = i ∙ √(x - a) and e^(i ∙ Θ) = cos Θ + i ∙ sin Θ
= -e^(i ∙ √(x - a)) ÷ (2 ∙ i ∙ √(x - a))
= -e^(i ∙ √(x - a)) ÷ (2 ∙ i ∙ √(x - a))
= i ∙ e^(i ∙ √(x - a)) ÷ (2 ∙ √(x - a))
= (i ∙ (cos √(x - a) + i ∙ sin √(x - a)) ÷ (2 ∙ √(x - a))
= (-sin √(x - a) + i ∙cos √(x - a)) ÷ (2 ∙ √(x - a))
Eddie
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