Let a curve have y = x + 1 / X-1, and the tangent at point (3,2) is perpendicular to the straight line ax + y + 1 = 0, then a is equal to? y=(x+1)/(x-1)=1+[2/(x-1)] y'=-2/(x-1)² When x = 3, y '= - 2 / (3-1) & # 178; = - 1 / 2 The tangent of the point is perpendicular to the line ax + y + 1 = 0 That is, their slope product is - 1 -1/2×(-a)=-1 The solution is a = - 2, -1/2×(-a)=-1 Why - A is not clear

Let a curve have y = x + 1 / X-1, and the tangent at point (3,2) is perpendicular to the straight line ax + y + 1 = 0, then a is equal to? y=(x+1)/(x-1)=1+[2/(x-1)] y'=-2/(x-1)² When x = 3, y '= - 2 / (3-1) & # 178; = - 1 / 2 The tangent of the point is perpendicular to the line ax + y + 1 = 0 That is, their slope product is - 1 -1/2×(-a)=-1 The solution is a = - 2, -1/2×(-a)=-1 Why - A is not clear

The straight line ax + y + 1 = 0 can be converted into oblique section: y = - AX-1, (- A is the slope of the straight line), so it is
-1 / 2 × (- a) = - 1, (slope product is - 1),

The curve y = ax & # 178; - ax + 1 is perpendicular to the straight line 2x + y + 1 = 0 at point (0,1), then a=

F '(x) = 2ax-a, f' (0) = - A, the slope of the tangent passing through (0,1) is - A, the slope of the straight line k = - 2. Because the tangent is perpendicular to the straight line, we know: (- a) (- 2) = - 1A = - 1 / 2. Or from the slope of the straight line k = - 2, the product of the tangent and the straight line perpendicular, the slope of the tangent is - 1 / 2, and the tangent equation is: y = 1 / 2x + 1 generation curve: 1 / 2x +

Find the included angle of the tangent of the curve y = 2-x ＾ 2 / 2 and y = 1 / 4x ＾ 3-2 at the intersection

First, the intersection point (2,0) can be obtained by combining the two
Then, the derivatives of the two formulas are obtained, and the rates at this point are K1 = - 2, K2 = 3;
The tangent angle is set to a
tanA=|(k1-k2)/(1+k1*k2)|=1
So the angle is 45 degrees

Calculation: (1) (- 7) - 9 - (- 3) + (- 5) (2) - 4.2 + 5.7-8.4 + 1
Calculation: (1) (- 7) - 9 - (- 3) + (- 5)
（2）-4.2+5.7-8.4+10

-7-9+3-5=-18

How to find the Taylor formula of arctanx at x = 0? Directly use the Taylor expansion? Or use the original five kinds of known Taylor formula?
How to find the n-order derivative of arctanx?

(arctan(x))'=1/(1+x^2)
This derivative can be expanded by the basic formula 1 / (1 + x)

Calculation (1) | + 2 2 / 3 | + | - 9 | (2) | - 3 / 4 | + | - 1 7 / 8|

（1）|+2 2/3|+|-9|
=2+2/3+9
=11+2/3
=11 and 2 / 3;
（2）|-3/4|+|-1 7/8|
=3/4+1+7/8
=1+13/8
=2+5/8;
=2 and 5 / 8;
I'm very glad to answer your questions. Skyhunter 002 will answer your questions
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Let the probability density function f (x) = {32 / (x + 4) ^ 3, x > 0, others} of continuous random variable x, and let the random variable y = x + 4, then find e (y)

We can integrate directly according to the definition of expectation, where p (x) & nbsp; = & nbsp; f (x) is the density function

How to calculate the determinant?

Solution: line I proposes I, I = 2,3,..., n
D = n!*
1 1 ... 1
1 2 ... 2^(n-1)
1 3 ... 3^(n-1)
... ... ...
1 n ... n^(n-1)
This is the Vandermonde determinant
D = n!(n-1)!(n-2)!...2!1!.

Let two random variables (x, y) be uniformly distributed in the region D, where d = {(x, y): | x | + | y | ≤ 1}
The probability density f (U) and f (V) of u and V?
The joint density is 1 / 2;
That is, when - 1 ≤ u ≤ 1, ∫ (d) 1 / 2dxdy (D: x + y ≤ U), how should we find it here?
Similarly, when - 1 ≤ V ≤ 1, how to find?
∫ (d) 1 / 2dxdy (D: x + y ≤ U), the answer to this step is (U + 1) / 2; when - 1 ≤ V ≤ 1, it is also this value; but I can't get this number. I don't know how to deal with the upper and lower limits of XY integral here.

The integral variable is 1 / 2. Do you have to integrate it? If you don't want to find the result, then you will be in the definite integral interval between y = u-x and y = - 1-x. (take the first one as an example). It's a bit troublesome. How simple is it in geometric sense. It's too troublesome for you. I mistook u just now. I just took it as the upper part. Sorry, the D area is a square

Calculation of determinant 2 2 2 2 2 … 2 2 2 2 3 … 2 2 …… …… 2 2 2 … 2 n

All lines minus line 2
Column 1 minus column 2
The determinant is transformed into upper triangular form
D = -1*2*1*2*...*(n-2) = -2*(n-2)!