1. To find the derivative of f(x) = sqrt(1-x^2), we can use the chain rule:
f'(x) = -x / sqrt(1-x^2)
2. To find the critical points of f(x) = 4x^3 + 3x^2 - 6x + 1, we need to find the values of x where f'(x) = 0 or f'(x) is undefined. First, we find the derivative:
f'(x) = 12x^2 + 6x - 6
Setting f'(x) = 0, we get:
12x^2 + 6x - 6 = 0
Simplifying, we get:
2x^2 + x - 1 = 0
Using the quadratic formula, we get:
x = (-1 ± sqrt(1 + 8)) / 4
x = -1 or x = 1/2
So, the critical points are x = -1 and x = 1/2.
3. To use the second derivative test to find the local minima and maxima of f(x) = 1 + 3x^2 - 2x^3, we need to find the critical points and the second derivative:
f'(x) = 6x^2 - 6x
Setting f'(x) = 0, we get:
6x^2 - 6x = 0
Simplifying, we get:
6x(x - 1) = 0
So, the critical points are x = 0 and x = 1.
f''(x) = 12x - 6
At x = 0, f''(0) = -6, so f(x) has a local maximum at x = 0.
At x = 1, f''(1) = 6, so f(x) has a local minimum at x = 1.