Find the differential of the root x power of E ∫ the root x power DX of E

Find the differential of the root x power of E ∫ the root x power DX of E

First question:
d［e^（√x）］＝［e^（√x）］d（√x）＝［e^（√x）］/（2√x）.
Second question:
Let √ x = t, then: x = T ^ 2, ‡ DX = 2tdt
∴∫［e^（√x）］dx
＝2∫t（e^t）dt＝2∫td（e^t）＝2te^t－2∫e^tdt＝2［e^（√x）］√x－2e^t＋C
＝2［e^（√x）］√x－2e^（√x）＋C

If the function f (x) = log (x2 ax + 1 / 2) has a minimum value, the value range of real number a is? The answer is (1, root 2) The function is loga (x2 ax + 1 / 2), which was missed just now

x2-ax+1/2=(x^2-ax+a^2/4)+(1/2-a^2/4)
The function u = x2 ax + 1 / 2 = (x ^ 2 ax + A ^ 2 / 4) + (1 / 2-A ^ 2 / 4) has a minimum value of 1 / 2-A ^ 2 / 4
The function f (x) = log (x2 ax + 1 / 2) has a minimum value,
1/2-a^2/4>0 a^2

If f (x) = log (a) {x ²- Ax + 1 / 2 、 has a minimum value, then the value range of real number a is the detailed solution

a> 1, G (x) = x ^ 2-ax + 1 / 2 = (x-a / 2) ^ 2 + 1 / 2-A ^ 2 / 4. If G (x) has a minimum value and is a positive number, f (x) has a minimum value. At this time, 1 / 2-A ^ 2 / 4 > 0, that is, 1

Find the monotone interval of the function f (x) = (1 / 2) * - x to the power of - 2x + 3

Negative infinity increases monotonically to negative two, and negative two decreases monotonically to positive

It is proved that the third power + 1 of the function FX = 2x is an increasing function on (negative infinity, positive infinity)

Set x1

It is proved that the function y = 2x to the fourth power increases on [0, + infinity]

Method 1: define the notation f (x) = y = 2x ^ 4, take X1 > x2 ≥ 0, then f (x1) - f (x2) = X1 ^ 4-x2 ^ 4 = (x1) ²+ x2 ²) (x1 ²- x2 ²)= (x1 ²+ x2 ²) (x1 + x2) (x1-x2) > 0 ‡ f (x1) > F (x2) ‡ from the definition, f (x) monotonically increases on [0, + ∞). Method 2: derivative method y