Curve y = SiNx sinx+cosx−1 2 at point m (π) The slope of tangent at 4,0) is () A. −1 Two B. 1 Two C. − Two Two D. Two Two

Curve y = SiNx sinx+cosx−1 2 at point m (π) The slope of tangent at 4,0) is () A. −1 Two B. 1 Two C. − Two Two D. Two Two

∵y＝sinx
sinx+cosx−1
Two
∴y'=cosx(sinx+cosx)−(cosx−sinx)sinx
(sinx+cosx)2
=1
(sinx+cosx)2
y'|x=π
4=1
(sinx+cosx)2|x=π
4=1
Two
Therefore, B

Curve y = SiNx sinx+cosx−1 2 at point m (π) The slope of tangent at 4,0) is () A. −1 Two B. 1 Two C. − Two Two D. Two Two

∵y＝sinx
sinx+cosx−1
Two
∴y'=cosx(sinx+cosx)−(cosx−sinx)sinx
(sinx+cosx)2
=1
(sinx+cosx)2
y'|x=π
4=1
(sinx+cosx)2|x=π
4=1
Two
Therefore, B

The point where the tangent slope of the sine curve y = SiNx is equal to 1 / 2 is____________

y=sinx,y‘=（sinx）’=cosX=1/2
X=2Kπ±π/3
When x = 2K π ± π / 3, y = sin (2k π ± π / 3) = ± √ 3 / 2
So the point where the slope of the tangent on the sine curve y = SiNx equals 1 / 2 is zero
(2k π + π / 3, √ 3 / 2) or (2k π - π / 3, √ 3 / 2)

Curve y = SiNx sinx+cosx−1 2 at point m (π) The slope of tangent at 4,0) is () A. −1 Two B. 1 Two C. − Two Two D. Two Two

∵y＝sinx
sinx+cosx−1
Two
∴y'=cosx(sinx+cosx)−(cosx−sinx)sinx
(sinx+cosx)2
=1
(sinx+cosx)2
y'|x=π
4=1
(sinx+cosx)2|x=π
4=1
Two
Therefore, B

The point where the slope of tangent on the sine curve y = SiNx (x ∈ (0,2 π)) is 1 / 2

The slope of tangent on the sine curve y = SiNx (x ∈ (0,2 π)) is equal to its derivative
Y = derivative of SiNx y '= cosx cosx = 1 / 2 x = 3 π / 4, where y = SIN3 π / 4 = (radical 3) / 2
This point is (3 π / 4, (radical 3) / 2)

Given the curve y = 5 times the root sign X. find the tangent equation parallel to the straight line y = 2X-4 on the curve; and find the tangent equation that passes through the point P (0,5) and is tangent to the curve

① Solution: let the tangent cross the point x0 on the curve y = 5x ^ 1 / 2
From y = 5x ^ 1 / 2, the slope of the tangent line is obtained
Y │ x = x0 = 5 / (2 pieces of x0)
The slope of the tangent parallel to the line y = 2X-4 is 2
5 / (2 root sign x0) = 2
X 0 = 25 / 16
Substituting into the curve y = 5x ^ 1 / 2, the
y0=25/4
Therefore, the tangent equation is obtained
y-25/4=2(x-25/16)
Namely:
16x-8y+25=0

(1 / 2) known curve y = f (x) = 5 (root x) find the tangent equation 2 crossing point (0,5) which is parallel to the straight line y = 2X-4 and tangent to the curve (1 / 2) given the curve y = f (x) = 5 (root x), find the tangent equation of 1 Curve parallel to the straight line y = 2X-4, and tangent to the curve through point (0,5)

Derivative function y '= 2.5 / √ x
∵ the tangent line is parallel to the line y = 2X-4
ν 2.5 / √ x = 2 ℅ x = 25 / 16, tangent point (25 / 16, 5 / 4)
The tangent equation y-25 / 4 = 2 (x-25 / 16)
If the point (0,5) is not on the curve, let the linear equation be y = KX + 5, and substitute it into the curve equation, and use △ = 0 to get k = 5 / 4
The linear equation y = 5x / 4 + 5

Given the curve y = 2 √ x + 1. (PS: 1 is not in the root sign.) find the tangent equation at which point of the curve is perpendicular to the straight line 2x + Y-3 = 0

y'=x^(-1/2),
Let the tangent line at P (x0, Y0) be perpendicular to the straight line, the straight line 2x + Y-3 = 0, and the slope K1 = - 2
Then the tangent slope K2 is its negative reciprocal, which is 1 / 2,
x0^(-1/2)=1/2,
x0=4,y0=5,
So the tangent of the curve at P (4,5) is perpendicular to the line 2x + Y-3 = 0

Curve y = (e ^ (2x)) multiplied by cos (3x) tangent at (0,1) and the distance between L is the root sign 5, and find the equation of L

Y = (e ^ (2x)) times cos (3x) y '= 2 (e ^ (2x)) times cos (3x) + (e ^ (2x)) times (- sin (3x)) * 3Y' | (x = 0) = 2. Here, the tangent Y-1 = 2X is 2x-y + 1 = 0. Let l equation 2x-y + k = 0. Let the distance between tangent and l be root 5: | k-1 | / under root (2 ^ 2 + 1 ^ 2) = root 5| k-1 | 5K = - 4 or 6

If we know that the equation of Y + X is perpendicular to the X + 2 curve at the point of the root of the curve If it exists, write out the coordinates of this point and find the tangent equation at the point. If not, give the reason

At the point of - 2x P, the tangent line of the equation is equal to the tangent line
The tangent equation can be assumed to be y = 1 / 2x + a
Suppose point P exists
Then the system of equations
y=2√x+1
There is and only one root
y=1/2x+a
The equation system is eliminated to obtain 1 / 4x ^ 2 + (A-4) x + A ^ 2-4 = 0
Since the above equation has and only one root, the left side of the equation can be completely formulated
Let the equation be (1 / 2x + b) ^ 2 = 0
∴b=a-4 b^2=a^2-4
∴a=5/2
The tangent equation at point P is y = 1 / 2x + 5 / 2
Therefore, the only root of the above equations is x = 3, y = 4
The hypothesis holds, that is, the point P exists
Point P coordinates are (3,4) tangent equation is y = 1 / 2x + 5 / 2