Differentiation (function) Find d ^ 2Y / DX ^ 2 of each function 1.y=cosx/(√x) 2.x^3-3xy+y^3=4 3.sin y +cos x=1 4.x^2/3+y^2/3=a^2/3 1.1/4x^-5/2[(3-4x^2)cosx +4xsinx] 2.10xy(x-y^2)^-3 3.(cos^2ycos x+sin^2xsin y)/cos^3y 4.(a^2/3)/(3x^4/3 y^1/3)

Differentiation (function) Find d ^ 2Y / DX ^ 2 of each function 1.y=cosx/(√x) 2.x^3-3xy+y^3=4 3.sin y +cos x=1 4.x^2/3+y^2/3=a^2/3 1.1/4x^-5/2[(3-4x^2)cosx +4xsinx] 2.10xy(x-y^2)^-3 3.(cos^2ycos x+sin^2xsin y)/cos^3y 4.(a^2/3)/(3x^4/3 y^1/3)

1.y'=-sinx/(√x)-cosx*(√x)³/2
y"=[-cosx/(√x)+sinx*(√x)³/2]
+[sinx*(√x)³/2+3cosx*/4(√x)^5]
=1/4x^-5/2[(3-4x^2)cosx +4xsinx]
2.x^3-3xy+y^3=4
The two sides are derived at the same time
3x²-3y-3xy'+3y²y'=0
That is X & sup2; - y-xy '+ Y & sup2; = 0 (1)
Again, both sides of the derivation at the same time
2x-y'-y'-xy”+2yy'=0
That is, 2x-2y '- XY "+ 2XY' = 0 (2)
Eliminate y 'from (1) (2)
We get y "= 10xy (X-Y ^ 2) ^ - 3
3.sin y +cos x=1
The two sides are derived at the same time
cosyy'-sinx=0 (1)
Again, both sides of the derivation at the same time
-siny(y')²+cosyy"-cosx=0 (2)
Eliminate y 'from (1) (2)
Get y "= (COS ^ 2ycos x + sin ^ 2xsin y) / cos ^ 3Y
4.x^2/3+y^2/3=a^2/3
The two sides are derived at the same time
2x/3+2yy'/3=0 （1）
Again, both sides of the derivation at the same time
2/3+2（y'）²/3+2yy“/3=0 （2)
Eliminate y 'from (1) (2)
Get y "= (a ^ 2 / 3) / (3x ^ 4 / 3, y ^ 1 / 3)

Find the differential of function y = 5 ^ (LN Tan x)! In my book, finding the differential is derivative multiplied by DX, but the result I got is wrong!

It is decomposed into y = 5 ^ u, u = LNv, v = TaNx by the derivative rule of composite function;
dy/dx=dy/du×du/dv×dv/dx
Using the derivation formula, dy / Du = 5 ^ u × LN5, Du / DV = 1 / V, DV / DX = (secx) ^ 2, we can put it into a function about X
The differential is dy = (dy / DX) DX

If a > 1, b > 1, then loga (b) + logb (a) ≥ 1___

Loga (b) and logb (a) are reciprocal and both are positive numbers. Then loga (b) + logb (a) ≥ 2 [[unclear, ask again; satisfied, please adopt! Good luck!]]

The definition field of function f (x) = LG (3-2x-x ^ 2) is_______

True number 3-2x-x & # 178; > 0
x²+2x-3=(x+3)(x-1)

1 and 2 / 7 fractional units are fractions. It contains several such fractional units. Adding several such fractional units is the smallest prime number

1 / 7.9, 12

Finding the marginal distribution by knowing the probability density of two-dimensional random variables

Let fxy (x, y) be the probability density function
The edge density function FX (x) = fxy (x, y) dy of X integrates from negative infinity to positive infinity (x is regarded as a constant when integrating)
The edge density function FY (y) = fxy (x, y) DX of Y is integrated from negative infinity to positive infinity (y is regarded as a constant when integrating)

Find the maximum value of the function y = cosx sin ^ x-cos2x + 7 / 4, and write the set 3Q of X when y takes the maximum value

When y = cosx - (SiNx) ^ 2 - (cosx) ^ 2 + (SiNx) ^ 2 + 7 / 4 = - [(cosx) ^ 2-cosx-7 / 4] = - [(cosx-1 / 2) ^ 2-2] = 2 - (cosx-1 / 2) ^ 2cosx belongs to [- 1,1] cosx = 1 / 2, y (max) = 2, X belongs to {x | x = (+, -) π / 3 + 2K π, k = 0, (+, -) 1, (+, -) 2.} cosx = - 1 / 2, y (min) = 1, X belongs to {x | x = (+, -)

If two-thirds of a is equal to three-quarters of B, then a is () of B

Then a = (b * 3 / 4) / (2 / 3) = (b * 3 / 4) * 3 / 2 = b * 9 / 8

The bottom length of trapezoid is two-thirds of the top length. The trapezoid is divided into a triangle and a parallelogram. What is the area of triangle in the original trapezoid?
If you can make a formula, make a formula,

If the length of the bottom of trapezoid is two-thirds of that of the top, the bottom of triangle is half of that of quadrilateral, and the height is equal
So triangle area / quadrilateral area = 1 / 4
So the triangle area accounts for 1 / (1 + 4) = 1 / 5 of the original trapezoid area

As shown in the figure, points a (m, M + 1) and B (M + 3, m-1) are all on the image of inverse scale function y = KX
(1) Through point B, make BC ⊥ X axis at C. on the image of inverse scale function, is there a point P between points a and B, so that the area of △ OAB is equal to the area of △ POC? If so, ask for the analytical expression of straight line Op
(2) Through point a, make am ⊥ Y axis in M, through point B, make BN ⊥ X axis in N, connect Mn. When point a and point B move on the image of inverse scale function (point a and B do not coincide), what is the position relationship between Mn and ab? Please prove your conclusion

(1) Substituting a (m, M + 1), B (M + 3, m-1) into y = K / X
M + 1 = K / m ^ 2 + M = K (1)
m-1=k/(m+3) m^2+2m-3=k (2)
The solution of simultaneous (1) (2) is m = 3, k = 12
So a (3,4) B (6,2) y = 12 / X
So let P (x ', 12 / X')
The area of △ OAB = (1 / 2) * 3 * 4 + (1 / 2) (4 + 2) * (6-3) - (1 / 2) * 6 * 2 = 12
Area of △ POC = (1 / 2) * 6 * (12 / X ') = 36 / X'
Then 36 / X '= 12 x' = 3
So p (3,4)
The analytic expression of OP is y = (4 / 3) X
(2) AB and Mn are parallel
Let the inverse scale function be y = K / X
So let a (a, K / a) B (B, K / b)
Then M (0, K / a) n (B, 0)
The slope of line AB k = (K / A-K / b) / (a-b) = - K / AB
The slope of line Mn K '= (K / A-0) / (0-B) = - K / AB
So k = k '
So abiimn
Get proof