Monday, November 8, 2021

Derivatives with Surprisingly Imaginary Results

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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