X / 2 = Y / 3 = Z / 4, find (XY + YZ + XZ) / (x ^ 2 + y ^ 2 + Z ^ 2)

X / 2 = Y / 3 = Z / 4, find (XY + YZ + XZ) / (x ^ 2 + y ^ 2 + Z ^ 2)

Let X / 2 = Y / 3 = Z / 4 = K
Then, x = 2K, y = 3k, z = 4K
（xy+yz+xz)/(x^2+y^2+z^2)
=(6k^2+12k^2+8k^2)/(4k^2+9k^2+16k^2)
=26k^2/29k^2
=26/29

How to find the partial derivative of root XY?
And the derivative of the whole function?

The partial derivative of (sqrt denotes square root) to X: if y is regarded as a constant, then sqrt [XY] = sqrt [y] * sqrt [x], where sqrt [y] is a constant, and then sqrt [x] = x ^ (1 / 2) is derived from X to get (1 / 2) x ^ (- 1 / 2) = 1 / (2sqrt [x]). So the partial derivative of sqrt [XY] to x = sqrt [y] * 1 / (2sqrt [x]) = sqrt [y] / (2sqrt [x])

How to find the partial derivative of Z = arctan (XY)

∂z/∂x=[1/(1+(xy)²)]*y=y/(1+x²y²)
∂z/∂y=[1/(1+(xy)²)]*x=x/(1+x²y²)

Partial derivatives of higher numbers with respect to implicit functions
Let e ^ z-xyz = 0, find (second order partial Z is more than partial x)
Fx=-yz,Fy=-xz,Fz=e^z-xy
yt xz
Partial Z to partial x = --- partial Z to partial y=------------
e^z-xy e^z-xy
z'(e^z-xy)-yz[(e^z) z'-xy]
Second order partial Z is better than partial X=----------------------------------
(e^z-xy)^2
2(y^2) z (e^z) -2(y^3) x z-(y^2)(z^2)(e^2)
=-----------------------------------------------------
(e^z-xy)^3
I would like to ask you that "second order partial Z is more partial X" is a division method as a whole, but I can't work out his result. What is the method used to deal with molecules? Just give me some ideas, and you don't need to use a formula,
”The molecule of partial Z is more than partial X "is YZ

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The problem of partial derivatives in higher numbers
How to find the partial derivative of function z = (x * x * y * y) / (X-Y) to independent variable x? It's written in the book
Taking y as a constant and deriving from X, we get
【2X*Y*Y（X-Y）-X*X*Y*Y】/(X-Y)(X-Y)
Under simplification, it is [x * x * y * y-2x * y * y * y] / (X-Y) (X-Y)
I want to ask how to get it step by step. My problem is that I don't understand "regarding y as a constant and deriving X"

It's all psychological. When you're on grass paper
Change the sign of partial derivative into the sign of derivative
Change y to C
Write again,
After that, replace it

Partial derivatives of higher numbers
The partial derivative of Z = lnsin (x-2y) There should be a process
It's useful to give an answer before 12 o'clock

The first is partial x, partial z = [1 / sin (x-2y)] cos (x-2y) partial X
Partial y, partial z = - 2 [1 / sin (x-2y)] cos (x-2y) partial x

A problem of partial derivative of higher number
Let r = root sign (x ^ 2 + y ^ 2 + Z ^ 2)
x. The second partial derivatives of Y and Z add up to 2 / R
I don't know how to prove it. I want to teach,

r = √(x²+y²+z²)∂r/∂x = 2x / 2√(x²+y²+z²) = x/r∂²r/∂x² = (1*r - x*x/r)/r² = (r²-x²)/r³ = (y²+z²)/r³...

The problem of higher numbers, the problem of partial derivatives
Is there a formula
f(x,y,z)'=f(x',y',z')

There is no such formula

Seeking partial derivative of higher number
Finding partial Z / partial x
z=yln(x+√(x²+y²))+3x-2y
Thank you for the details. X & sup2; is the square of X

z = yln[x+√(x²+y²)]+3x-2y
∂z/∂x = {y/[x+√(x²+y²)]}{1 + x/√(x²+y²)} + 3
= y[√(x²+y²)+x]/[x+√(x²+y²)][√(x²+y²)] + 3
= y/[√(x²+y²)] + 3

Let z = f (x, y) and x = t + Sint, y = T ^ 2, f (x, y) be differentiable, then DZ / DT
Let z = f (x, y) and x = t + Sint, y = T ^ 2, f (x, y) be differentiable, and find DZ / dt

The reciprocal of F to x multiplied by (1 + cost) plus the reciprocal of F to y multiplied by 2T