How to find the partial derivative of Z = arctan (x / y)

How to find the partial derivative of Z = arctan (x / y)

On the partial derivative of X: &; Z / &; X = 1 / y [1 + (x / y) ^ 2]
On the partial derivative of Y: &; Z / &; y = - X / y ^ 2 [1 + (x / y) ^ 2]

Finding partial derivative z = arctan (X-Y ^ 2)

This paper is an arctan (X-Y & \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\# 8706; y = {1 / [1 + (X-Y & # 178;) &# 178;]} × (X-Y & # 178;)

On the first partial derivative of Z = arctan (Y / x)
Z = the first partial derivative of arctan (Y / x)
∂z/∂x= {1/[1+(y/x)²]}·(-y/x²)
= -y/(x²+y²)
∂z/∂y= {1/[1+(y/x)²]}/x
= x/(x²+y²)
I want to ask, isn't the derivative formula of arctanx 1 / (1 + X & # 178;)? How do you get the part of (- Y / X & # 178;) on the right side of the braces in the formula of arctanx = {1 / [1 + (Y / X) & # 178;]} (- Y / X & # 178;),

∂ Z / & # 8706; X = {1 / [1 + (Y / x) & # 178;]} (Y / x) = {1 / [1 + (Y / x) & # 178;]} (- Y / X & # 178;) (this is the derivation of a composite function, that is, to derive X in (Y / x), that is, (Y / x) = - Y / X & # 178;)

Finding the first partial derivative: z = arctan √ (x ^ y)

z'(x)=1/[1+(x^y)] * 1/2√(x^y) * yx^(y-1) =yx^(y-1) / {2√(x^y)[1+(x^y)] }
z'(y)=1/[1+(x^y)] * 1/2√(x^y) * lnx *x^y=(x^y) *lnx / {2√(x^y)[1+(x^y)] }

Find the second partial derivatives of the following functions: 1) z = x ^ 4 + 3 * x ^ 2 * y + y ^ 3 2) z = XLN (x + y)
Find the second partial derivatives of the following functions
1）z=x^4+3*x^2*y+y^3
2）z=xln(x+y)

z=x^4+3x²y+y³
∂z/∂x = 4x³+6xy
∂z/∂y = 3x²+3y²
∂²z/∂x² = 12x²+6y
∂²z/∂x∂y = 6x
∂²z/∂y² = 6y
----------------------------
z=xln(x+y)
∂z/∂x = ln(x+y) + x/(x+y)
∂z/∂y = x/(x+y)
∂²z/∂x² = 1/(x+y) + [(x+y)-x]/(x+y)² = 1/(x+y) + y/(x+y)² = (x+2y)/(x+y)²
∂²z/∂x∂y = 1/(x+y) - x/(x+y)² = y/(x+y)²
∂²z/∂y² = - x/(x+y)²

Find the second order partial derivatives of the function z = x ^ 4 + y ^ 4-4xy э ^ 2 Z / x ^ 2, э ^ 2 Z / y ^ 2, э ^ 2 Z / x ^ y

э^2 z/эx^2=12x^2;
э^2 z/эy^2=12y^2;
э^2 z/эxэy=-4.
эz/эx=4x^3-4y;эz/эy=4y^3-4x;
So:
э^2 z/эx^2=э(эz/эx)/эx=э(4x^3-4y)/эx=4*3x^2=12x^2;
э^2 z/эy^2=э(эz/эy)/эy=э(4y^3-4x)/эy=4*3y^2=12y^2;
э^2 z/эxэy=э(эz/эy)/эx=э(4y^3-4x)/эx=-4.

Let the implicit function determined by the equation x + 2Y + Z = e ^ (X-Y-Z) be z = Z (x, y), and find d ^ 2Z / DX ^ 2

X + 2Y + Z = e ^ (X-Y-Z) two sides find partial derivatives of X. note that z = Z (x, y)
1+z'=e^(x-y-z)*(1-z')...(1)
Then the partial derivative of X is obtained
z"=e^(x-y-z)(1-z')^2-z"e^(x-y-z)...(2)
(1) The solution of Z 'can be substituted into (2)
We can get Z "= D ^ 2Z / DX ^ 2 =

If X & # 178; - ax + 1 > 0 is constant for X ∈ [0.5,3], find the value range of real number a

It is obvious that this function has the axis of symmetry x = A / 2
If X & # 178; - ax + 1 > 0 holds for X ∈ [0.5,3], it is required that
1,f(0.5)>0,f(3)>0,a/23
So it can be A0 = > A0 = > A = 0.5 = > a > = 1 A / 2A

F (x) = ax & # 178; - LNX, for any x ∈ (0, e,] f (x) ≥ 3, find the value range of real number a

F (x) = ax & # 178; - LNX, for any x ∈ (0, e,] f (x) ≥ 3, find the value range of real number a
Domain: x > 0;
From F '(x) = 2aX - (1 / x) = (2aX & # 178; - 1) / x, we can see that when a ≤ 0, for any x > 0, there is f' (x) 0, which is contradictory to the specified condition a ≤ 0, so this situation is not suitable
Existence
When a > 0, from 2aX & # 178; - 1 = 0, we get X & # 178; = 1 / (2a), then we get the stationary point x = √ (1 / 2a); when x0, so x = √ (1 / 2a) is the minimum point, the minimum value of F (x) = f [√ (1 / 2a)] = a (1 / 2a) - ln [√ (1 / 2a)] = 1 / 2 - (1 / 2) [- ln (2a)]
=1 / 2 + (1 / 2) ln (2a) ≥ 3, that is 1 + ln (2a) ≥ 6, ln (2a) ≥ 5, LN2 + LNA ≥ 5, LNA ≥ 5-ln2, so a ≥ e ^ (5-ln2)
=(1 / 2) e ^ 5, that is, a ∈ [(1 / 2) e ^ 5, + ∞), which is the value range of A

X & # 178; + 3 ≥ - ax-a is constant for any x [- 2,2]. Find the value range of real number a

Take x = - 2, x = 2, x = 0 to calculate the value of a, and get - 7 / 3