What is the fourth derivative of

f(x)= e^-x^2.

I got f''(x)=-2e^-x^2 + 4x^2e^-x^2

i need some help on find f''''(x)

You have
f(x) = e-x2 then
f'(x) = -2x*e-x2 = -2x*f(x)
f"(x) = (4x2 - 2))e-x2 = f(x)*(4x2 - 2))
f"'(x)= f'(x)*(4x2 - 2)) + f(x)(8x) = -2x*f(x)*(4x2 - 2)) =
f(x)(-8x3 + 12x)
f""(x) = f'(x)*(-8x3 + 12x) + f(x)(-242 + 12)=
-2x*f(x)(-8x3 + 12x) + f(x)(-242 + 12) =
f(x)(16x4-48x + 12)

Be sure to verify my algebra.

To find the fourth derivative of f(x) = e^(-x^2), we can start by finding the first derivative, then the second derivative, then the third derivative, and finally the fourth derivative.

1. First derivative (f'(x)):
Using the chain rule, we differentiate e^(-x^2) with respect to x and multiply it by the derivative of the exponent (-x^2):
f'(x) = d(e^(-x^2))/dx = -2x * e^(-x^2)

2. Second derivative (f''(x)):
Differentiate f'(x) with respect to x:
f''(x) = d(-2x * e^(-x^2))/dx = -2 * e^(-x^2) + (-2x) * (-2x * e^(-x^2))
= -2 * e^(-x^2) + 4x^2 * e^(-x^2)
So, the second derivative is f''(x) = -2 * e^(-x^2) + 4x^2 * e^(-x^2)

3. Third derivative (f'''(x)):
Differentiate f''(x) with respect to x:
f'''(x) = d(-2 * e^(-x^2) + 4x^2 * e^(-x^2))/dx
= (-2x * e^(-x^2))*(4x^2 - 2) + (-2 * e^(-x^2))*(8x)
= -2x * e^(-x^2)*(4x^2 - 2) + (-2 * e^(-x^2))*(8x)
Simplifying this expression gives: f'''(x) = -2x * e^(-x^2)*(4x^2 - 2) + (-2 * e^(-x^2))*(8x)

4. Fourth derivative (f''''(x)):
Differentiate f'''(x) with respect to x:
f''''(x) = d(-2x * e^(-x^2)*(4x^2 - 2) + (-2 * e^(-x^2))*(8x))/dx
= (-2 * e^(-x^2)*(4x^2 - 2))*(-8x) + (-2x * e^(-x^2))*(16x^3 - 24x)
= (-8x * e^(-x^2))*(4x^2 - 2) + (-32x^4 * e^(-x^2)) + 48x^2 * e^(-x^2) - 16x * e^(-x^2)
Simplifying this expression gives: f''''(x) = (-8x * e^(-x^2))*(4x^2 - 2) + (-32x^4 * e^(-x^2)) + 48x^2 * e^(-x^2) - 16x * e^(-x^2)

So, the fourth derivative of f(x) = e^(-x^2) is f''''(x) = (-8x * e^(-x^2))*(4x^2 - 2) + (-32x^4 * e^(-x^2)) + 48x^2 * e^(-x^2) - 16x * e^(-x^2).