Suppose that f(0) = 0 and f'(0) = 2 and let g(x) = f(-xf(f(x))). The value of g'(0) is equal to ........

Let f : (-1, 1) to R be a differentiable function with f(0) = -1 and f'(0) = 1 . Let g(x) = [f(2f(x) + 2)]^2 . Then g'(0) is equal to

f(x) = x^(2)+xg'(1) +g''(2) and g(x) =f(1)x^2+xf'(x)+f''(x) The value of g(0) is

If 2f(x) = f'(x) and f(0) = 3 then f(2) =

If f(x) = |x - 2| and g(x) = f(f(x)) , then sum_(r = 0)^(3) g'(2r - 1) is equal to

f(x) = x^(2)+xg'(1) +g''(2) and g(x) =f(1)x^2+xf'(x)+f''(x) The value of f(3) is

Let f(x) be continuous on [0, 4] , differentiable on (0, 4) , f(0) = 4 and f(4) = -2 . If g(x) =(f(x))/(x + 2) , then the value of g'(c ) for some Lagrange's constant c in (0 , 4) is

If f(0) = 0 , f '(0) = 2 , then the derivative of y = f(f(f(f(x)))) at x = 0 in