How do you do these questions?

Step-by-step explanation:

(a)

g(x) = d/dx [2/√π ∫₀ˣ e^(-t²) dt]

g(x) = 2/√π e^(-x²)

g(-x) = 2/√π e^(-(-x)²) = 2/√π e^(-x²)

g(-x) = g(x)

Therefore, g(x) is even.

(b)

erf(x) = 2/√π ∫₀ˣ e^(-t²) dt

erf(x) = ∫₀ˣ g(t) dt

erf(-x) = ∫₀⁻ˣ g(t) dt

erf(-x) = -∫₋ₓ⁰ g(t) dt

Since g(t) is even, it is symmetrical about x=0.

erf(-x) = -∫₀ˣ g(t) dt

erf(-x) = -erf(x)

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