The derivative of y = sin ^ n * x + sin NX! The derivative of y = ln (LN (1 + X / x))

The derivative of y = sin ^ n * x + sin NX! The derivative of y = ln (LN (1 + X / x))

y'=nsin^(n-1)*x+cosnx*n
(1/x)'=-1/x^2;
y'=[ln(ln(1+x/x))]'=[1/ln(1+x/x)]*[ln(1+x/x]'=[1/ln(1+x/x)]*x/(1+x)*(1+x/x)'=1/ln(1+x/x)]*x/(1+x)*1/x^2

Derivative of y = [sin radical (1-2x)] ^ 2

Let y = u ^ 2, u = SINV, v = √ W, w = 1-2x
y'=2u.cosv.1/(2√w).(-2)= -4sin[√(1-2x)].cos√(1-2x).1/[2√(1-2x)]

4. If one of the two roots of equation x2 + 2aX + A + 1 = 0 is larger than 2 and the other is smaller than 2, then the value range of real number a is

First of all, it shows that the quadratic equation with one variable has two unequal real roots, then the discriminant B ^ 2-4ac = a ^ 2-a-1 > 0 with roots is solved to a > (1 + sqrt5) / 2 or a

The product of a number and its reciprocal plus a, the result is 16 / 5, what is the reciprocal of a

The product of a number and its reciprocal = 1
1+a=16/5
a=11/5
1/a=5/11

Find the equation of a line perpendicular to the line 3x-4y-3 = 0 and passing through the origin

The slope of the line perpendicular to the line 3x-4y-3 = 0 and passing through the origin is k = - 1 / (3 / 4) = - 4 / 3
So: the equation is y = - 4 / 3x

The function f (x) = x ^ 2-x-m has a zero point in the interval (- 1,0), so we can find the value range of M

f(x)=x^2-x-m
Axis of symmetry x = 1 / 2, opening up!
There is a zero point in the interval (- 1,0)
Satisfy f (0) * f (- 1) 0
That is: 1 + 4m > 0 m > - 1 / 4
(-m)*(2-m)

If the intersection point of parabola y = x & # 178; + 1 and straight line y = x + 3 is a, B, and the vertex of parabola is m, then the area of △ AMB is no graph

First of all, there are two functions to establish the equations and get two points, (2,5), (- 1,2); obviously, the vertex of the parabola is (0,1),
In the plane rectangular coordinate system, draw three points, connect the triangle, you can get the area s = 3;

What is the symmetry axis of the equation y = ax square + BX + C passing through point (1, - 4) (6, - 4)?

The X coordinates of coordinates (1, - 4) and (6, - 4) are 1 and 6 respectively, and the Y coordinates are all - 4
x1=1,x2=6
x=(x1+x2)/2=3.5
The axis of symmetry is: x = 3.5

The sum of two prime numbers is 39. How much is the product of these two prime numbers?

The prime numbers from 1 to 39 are: 2,3,5,7,11,13,17,19,23,29,31,37,39. If the sum of the two prime numbers is 39, then the two prime numbers are 2 and 37, then their product is 2 * 37 = 74

Using formula method to solve equation: x ^ 2-3ax + (2a-ab-b ^ 2) = 0

x^2-3ax+(2a^2-ab-b^2)=0
x²-3ax+(2a+b)(a-b)=0
[x-(2a+b)][x-(a-b)]=0
∴x=2a+b x=a-b