The derivative of the function y = x (x + 1) at x = 2 is

The derivative of the function y = x (x + 1) at x = 2 is

First, we derive y = x (x + 1)
y=x^2+x
y'=2x+1
Evaluation: y '(x = 2) = 5;
We can also use the definition of derivative!

Given that f (x) = AX3 + 3x2 + 2, the tangent slope of Nordic curve f (x) at x = - 1 is 4. It is necessary to find the value of A

This is a high school topic, if the topic is f (x) = AX3 + 3x2 + 2
The derivation of the equation f (x) is obtained
F(X)=3aX2 +6X
Because the tangent slope of the curve f (x) at is 4
So substitute x = - 1
4=3a（-1）2 +6（-1）
We get a = 10 / 3

Let f (x) = xlnx, if f ′ (x0) = 2, then x0=______ ．

If f (x) = xlnx  f '(x) = LNX + 1, then f' (x0) = lnx0 + 1 = 2, the solution is: x0 = e, so the answer is: E

It is known that the equation of a circle is (x-1) ∧ 2 + (Y-1) ∧ 2 = 1. The tangent equation of a point P (3,2) and a circle is obtained

Let the tangent equation be Y-2 = K (x-3), that is, kx-y-3k + 2 = 0, the distance from the center of the circle (1,1) to the straight line be: 1, that is, | k-1-3k + 2 | / root 2 = 1 square, then: (- 2K + 1) ^ 2 = 24K ^ 2-4k + 1 = 44K ^ 2-4k-3 = 0 (2k-3) (2k + 1) = 0k1 = 3 / 2, K2 = - 1 / 2, that is, the tangent equation is y = 3 / 2 (x-3) + 2 or y = - 1 / 2 (x-3) + 2

The equation of the tangent line passing through a ^ 2 (y = 25)

Let the tangent point be p (x ', y') ∵ a (5,15) ┌ (x '- 0) ^ 2 + (y' - 0) ^ 2 = 5 ^ 2 └ (5-0) (x '- 0) + (15-0) (y' - 0) and get the solution ┌ X1 = 5 ┌ x2 = - 4 └ Y1 = 0 └ y2 = 3 └ the two tangent points are: (5,0) and (- 4,3) and ∵ a (5,15) ∵ the tangent equation is x = 5 or 3y-4x-25 = 0, thank you ~

Solve the tangent equation of (5,15) direction circle x ^ 2 + y ^ 2 = 25
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The tangent equation of circle x ^ 2 + y ^ 2 = R ^ 2 is xX0 + yy0 = R ^ 2, and (x0, Y0) is the tangent point
The tangent equation is 5x + 15y = 25
That is, x + 3y-5 = 0

The tangent equation of the circle x ^ 2 + y ^ 2 = 4 passing through point a (2,4) is

First of all, draw pictures!
There are two
1. In the vertical direction passing through the point, i.e. x = 2
2. Make another tangent line, connect the point and the center of the circle, connect the center of the circle and the tangent point, get the right triangle, easy to get K

Find the tangent equation of the circle x 2 + y 2 = 4 from point a (2,4)

Obviously, x = 2 is one of the tangents; let y-4 = K (X-2), that is, kx-y + 4-2k = 0, and the distance from the center of the circle (0, 0) to the tangent is equal to the radius, then | 4 − 2K | K2 + 1 = 2, K = 34, 3x − 4Y + 10 = 0, and the tangent equation of the circle is x = 2, or 3x-4y + 10 = 0

How to find the tangent equation of P (2,4) through a point outside the circle (x-1) ∧ 2 + (Y-2) ∧ 2 = 1?

1, when the slope does not exist, x = 2 holds. 2, when the slope exists, y-4 = K (X-2). Using the distance from the center of the circle (1,2) to the straight line = radius, K. | 2-4 + K | / radical (1 + K ^ 2) = 1, k = 3 / 4, so the tangent equation is x = 2 or y-4 = 3 / 4 (X-2)

If passing through the two tangents of the leading circle at a point m (4, - 1) outside the circle x ^ 2 + y ^ 2 = 4, the linear equation passing through the two tangents is 4x-y-4 = 0

[Note: there is a way, I wonder if you can accept it? Connect the point m (4, - 1) and the origin o (0,0). The circle with the diameter of the line segment OM is: X & sup2; - 4x + Y & sup2; + y = 0. The equation is subtracted from the equation x & sup2; + Y & sup2; = 4, that is, the linear equation where the tangent chord is: 4x-y-4 = 0