What is the derivative of F (x) = ex / x

What is the derivative of F (x) = ex / x

f(x)=[e^x]/(x)
f'(x)=[(e^x)' × (x)－(e^x) × (x)']/(x ²)
=[xe^x－e^x]/(x ²)

Given the function f (x) = MX2 + lnx-2x, if M = - 4, find the maximum value of function f (x)

The first derivative of F (x) = - 4x ^ 2 - 2x + LNX is: F '(x) = - 8x + 2 + (1 / x) to make it 0, that is, 8x ^ 2 - 2x - 1 = 0, the solution is x = - 1 / 4 (rounded, ∵ true number > 0) or 1 / 2, and the maximum value of F (x) is obtained by substituting f (1 / 2) = - 1 - 1 + ln (1 / 2) = - 2 - LN2

If f (LNX) = 3x + 4, the expression of F (x) is __

Let t = LNX, then x = et,
Therefore, from F (LNX) = 3x + 4, f (T) = 3et + 4
That is, f (x) = 3EX + 4
So the answer is: F (x) = 3EX + 4

F (x) = 1 / 2 x ^ 2 - 3x + LNX, find the maximum value

Take the derivative of F (x), f '(x) = x-3 + 1 / x, Let f' (x) = 0, (it is known that the definition field of F (x) is x > 0), and the solution is X1 = (3-radical 5) / 2 or x2 = (3 + radical 5) / 2
F '(x) > 0 when x belongs to (0, x1) and (X2, positive infinity), f' (x) when x belongs to (x1, x2)

Given the function f (x) = (x2-3x + 2) LNX + 2008x-2009, the equation f (x) = 0 must have a real root () in which of the following ranges A. （0，1） B. （1，2） C. （2，3） D. （2，4）

∵ function f (x) is a continuous function, and f (1) = - 1 ＜ 0, f (2) = 4016-2009 = 2007 ＞ 0,
Therefore, the zero point of function f (x) is on (1,2),
So the answer is B

F derivative (LNX) = 1 + X, then f (x) = (1) x + e ^ x + C (2) e ^ x + e ^ 2x / 2

F derivative (LNX) = 1 + X
Let LNX = t
x=e^t
The original formula is changed to:
f'(t)=1+e^t
therefore
f(t)=t+e^t+c
f(x)=x+e^x+c
Select (1)