Ask a math problem about derivative. We know that f '(x) is the derivative of F (x). On the interval [0, positive infinity], f' (x) > 0 It is known that f '(x) is the derivative of F (x). In the interval [0, positive infinity], f' (x) > 0, and even function f (x) satisfies f (2x-1)

（1/3,2/3）
Because it is an even function, the two sides of the function are symmetric and should be a parabola. So 2x-1 > - 1 / 3 and 2x

If f (x) = f '(1) / x + 4x, then f' (1) =?
If f (x) = f '(1) / x + 4x, then f' (1) =?

F '(1) is a constant
f’(x)=-f’(1)/x^2+4
If x = 1, then f '(1) = - f' (1) / 1 + 4
f’(1)=2

If the function f (x) = (x ^ 2 + 2x + a) / X is monotonically increasing on [1, + ∞), then the value range of a is given

f'=(2x+2)*x-(x^2+2x+a)/x^2
=(x^2-a)/x^2
Monotonically increasing on [1, + ∞),
f'>=0
x^2-a>=0
x^2>=a
a

The relationship between the geometric meaning of the average rate of change of a function on an interval and its derivative

The derivative is positive, the function is monotone increasing, the function is negative, the function is monotone decreasing

Find the increasing and decreasing intervals of derivative of function y = x / 1 + X

y'=(x+1-x)/(x+1)^2=1/(1+x)^2>0
Because the domain of definition is x ≠ - 1, there is no decreasing interval, and the increasing interval is
(- ∞, - 1) and (- 1, + ∞)

F (x) = (X-5) / (x ^ 2-5x + 6) find the n-th derivative of this function

Obviously
f(x)=(x-5)/(x^2-5x+6)= -2/(x-3)+3/(x-2)
Then the derivative of order n is obtained
fn(x)= -2*(-1)^n *n!*(x-3)^(-n-1) +3*(-1)^n *n!*(x-2)^(-n-1)

If the derivative of function y = f (x) exists at x = 0, and f (x) = f (- x), find the value of F '(0)

f(x) = f(-x)f'(0-)= lim(y->0-) { [f(y) - f(0)]/y } = lim(y->0-) { [f(-y) - f(0)] / (y) }= -lim(y-> 0-) { [f(-y) - f(0)]/ (-y) }= -lim(-y-> 0+){ [f(-y) - f(0)]/ (-y) } = - f'(0+)f'(0) => f'(0+) = f'(0-...

Does the derivative of y = f (x) = x ^ (1 / 3) exist at 0?

y=x^(1/3)
y'=(1/3)*x^(-2/3)
=1/[3x^(3/2)]
And the derivative at x = 0 is
y'|(x=0)
=1/[3*0^(3/2)]
=1 / 0, (fraction meaningless)
The derivative of y = f (x) at x = 0 does not exist

F (x + y = f (x) f (y), and the derivative of F (0) exists. It is proved that f '(x) = f (x) f' (0)

Because: F (0 + 0) = f (0) * f (0) gets: F (0) = [f (0)] ^ 2 gets: F (0) = 0, or F (0) = 1, if f (0) = 0, then for any x, there is: F (x) = f (x + 0) = f (x) * f (0) = 0, so for any X: F '(x) = 0 proposition holds. If f (0) = 1, then: [f (x + H) - f (x)] / h = [f (x) * f (H) - f (x)] / h = f (x) * [f (H) - 1] / h = f (x) * [f (H

Derivative of y = x ^ 4-3x ^ 2 + 2x + e

4x^3-6x+2

Ask a math problem about derivative. We know that f '(x) is the derivative of F (x). On the interval [0, positive infinity], f' (x) > 0 It is known that f '(x) is the derivative of F (x). In the interval [0, positive infinity], f' (x) > 0, and even function f (x) satisfies f (2x-1)