erf(x)= 2/√pi ∫0 to x e^(-t^2) dt
a) Find erf (0)
For this I just imputed o to find the value and I got e^0=1 so 1(2/√pi ) to be my answer. But I was marked wrong. Then I thought that the function after the integral is the anti derivative so I might need to take the derivative of which I got a negative of my previous answer which was also wrong. I have been looking through my notes and the section in the book but I just don't understand this.
b) find d/dx(erf(x^3))
This is where I thought that the derivative is just a product. So I took the e^(-t^2)+(-x e^(-t^2)) and then substituted the t's for x^3 and got e^(-x^6)+(-x e^(-x^6)) but that was marked wrong. What should I do?
The error function, erf(x), is defined by the following?
2017-04-27 11:04 am
回答 (2)
2017-04-27 11:45 am
✔ 最佳答案
erf(x)= 2/√π [0,x] = ∫ e^(-t^2) dterf(0) = [0,0] ∫ ..... = 0
d/dx erf(x³)?
let u = x³
erf(u) = 2/√π [0,u] ∫ e^(-t²) dt
d/du erf(u) = 2/√π e(-u²) by FTC
using the chain rule
d/dx erf(u) = d/du erf(u) * du/dx
= 2/√π e^(-u²) du/dx
= 2/√π e^(-x⁶) * 3x²
= 6x²/√π e^(-x⁶)
2017-04-27 12:13 pm
a) erf(0)= (2/√π) ∫(t = 0 to 0) e^(-t^2) dt = 0.
b) (d/dx) erf(x^3)
= (d/dx) [(2/√π) ∫(t = 0 to x^3) e^(-t^2) dt]
= e^(-(x^3)^2) * (d/dx) x^3, by Fund. Thm of Calculus and the Chain Rule
= 3x^2 e^(-x^6).
I hope this helps!
b) (d/dx) erf(x^3)
= (d/dx) [(2/√π) ∫(t = 0 to x^3) e^(-t^2) dt]
= e^(-(x^3)^2) * (d/dx) x^3, by Fund. Thm of Calculus and the Chain Rule
= 3x^2 e^(-x^6).
I hope this helps!
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