Find the derivative of the implicit function determined by the equation y = Xe ^ y + 1? y=x e^y +1

Find the derivative of the implicit function determined by the equation y = Xe ^ y + 1? y=x e^y +1

Two sides of X are derived
y'=e^y+xy'e^y
Get: y '= e ^ y / (1-xe ^ y)

Urgent! Find the second derivative d ^ 2Y / DX ^ 2 of the function y determined by the following parametric equation
Find the second derivative d ^ 2Y / DX ^ 2 of the function y determined by the following parametric equation
1.x=2t-t^2, y=3t-t^3
2.x=f'(t), y=tf'(t)-f(t) (f''(t)≠0)

The first question is 3 / (4-4t)
The second question is 1 / F "(T)

When m is in what range, the intersection of y = 3x + M-1 and y = 2x-3m + 2 is in the third quadrant

The intersection point x = 3-2m can be calculated by linear equation
y=8-5m
The intersection point should be in the third m quadrant. X

①.(2x+1)^2-(7-x)^2=0 ②.3x(x+1)=2x+2

① (2x + 1) ^ 2 - (7-x) ^ 2 = 0 [(2x + 1) - (7-x)] [(2x + 1) + (7-x)] = 0 (2x + 1-7 + x) (2x + 1 + 7-x) = 0 (3x-6) (x + 8) = 0 (X-2) (x + 8) = 0x = 2 or x = - 8 ②. 3x (x + 1) = 2x + 23x ^ 2 + 3x = 2x + 23x ^ 2 + X-2 = 0 (3x-2) (x + 1) = 0x = 2 / 3 or x = - 1

Calculate by simple algorithm: (3.4 * 4.8 * 9.5) divide by (1.9 * 1.7 * 2.4)

=(3.4÷1.7)*(4.8÷2.4)*(9.5÷1.9)
=2*2*5
=20

If the function f (x) defined on R satisfies f (0) = 0, f (x) + F (1-x) = 1, f (x / 5) = 1 / 2F (x), then f (1 / 2008)=

Let x = 1 / 2F (1 / 2) + F (1-1 / 2) = 12 * f (1 / 2) = 1F (1 / 2) = 1 / 2F (x / 5) = 1 / 2F (x), so f (1 / 10) = f [(1 / 2) / 5] = 1 / 2 * f (1 / 2) = 1 / 4f (1 / 50) = f [(1 / 10) / 5] = 1 / 2 * f (1 / 10) = 1 / 8, repeat several times f (1 / 1250) = 1 / 32, let x = 1F (1) + F (1-1) = 1F (1) = 1-f (0) = 1F (x / 5) = 1 / 2F (x) F

The absolute value of minus 1 and 1 / 4 minus 2 and 1 / 3 minus 1 and 1 / 2

|-5/4-(-8/3)|-(-3/2)
=|-5 / 4 x 8 / 3 | x 3 / 2
=17 / 12 x 3 / 2
=35/12

Finding the domain of definition and monotone interval of logarithmic function
Find the domain of the function y = LG (x ^ 2 + 1 / 4 * X-5) and write its monotone interval

Domain x ^ 2 + 1 / 4 * X-5 > 0
4X ^ 2 + x-20 > 0 is defined as x > (- 1 + √ 321) / 8 or X (- 1 + √ 321) / 8 increasing interval
x

Change (- 20) - 9 - (- 8) + 10 into an expression without brackets and absolute value sign

To get rid of the brackets, we have to:
(-20)-9-(-8)+10
=-20-9+8+10
=-11 .
Replace the brackets with the sign of absolute value
/-20/-9-/-8/+10
=20-9-8+10
=13

If x > 0, f (x) = x & # 178; + LG (x + 1), then

Let x0, so f (- x) = (- x) 2 + LG (- x + 1), and because f (- x) = - f (x), so f (x) = - f (- x) = - x2 LG (- x + 1). [x]