The third power of Z = x is multiplied by the Y power of E + (x-1) arctanx, and the partial derivative of X is obtained

The third power of Z = x is multiplied by the Y power of E + (x-1) arctanx, and the partial derivative of X is obtained

Zx=3e^yx^2+arctan(y/x)-(x-1)(1/[(y/x)^2+1])*yx^(-2)

The derivative of X / y = ln (XY)

X/Y=lnX+lnY
YlnX+YlnY=X
Y'lnX+Y/X+Y'lnY+YY'/Y=1
Y'(lnX+lnY+1)=1-Y/X
Y' = (1-Y/X)/(lnX+lnY+1)

What is the partial derivative of Ln (XY),

How to subtract the differential of xy = 0 from the XY power of E?

e^(xy)-xy=0
This function is called implicit function
e^(xy)=xy
Take logarithm on both sides
xy=ln(xy)
Derivative on both sides
y+x*dy/dx=1/(xy)*[y+x*dy/dx]=1/x+1/y*dy/dx
(x-1/y)*dy/dx=1/x-y
dy/dx=[1/x-y]/(x-1/y)=(1-xy)/x*y/(xy-1)=-y/x
[question]: how to subtract the differential of xy = 0 from the XY power of E?
[answer]: differential dy = dy / DX * DX = - Y / X * DX

A problem of higher numbers, differential equation, the answer is y = (x / 3) + (C / x ^ 2)
x^2dy+(2xy-x^2)dx=0

X ^ 2dy + (2xy-x ^ 2) DX = 0 divided by x ^ 2 (2Y / x-1) DX + dy = 0, let Y / x = V dy = XdV + VDX (V ^ 2-1) DX + XdV + VDX = 0 (3v-1) DX + XdV = 0 DX / x = DV / (1-3v) integral on both sides: ln | x | = - 1 / 3ln | 1-3v | ln | x | = - 1 / 3ln | 1-3y / X | x ^ (- 2) = x-3y, so the final result is

Higher number: a sufficient condition for f (x, y) to be differentiable at a point is
(a) All second partial derivatives of F (x, y) are continuous
(b) , f (x, y) continuous
(c) All first partial derivatives of F (x, y) are continuous
(d) F (x, y) is continuous and continuous partial derivatives for X, y exist

All the first partial derivatives of C f (x, y) are continuous

Finding the differential K (x) = (the third power of 2x + 1) (the fourth power of X - 2x)
RT,

Decompose K (x) = (2x ^ 3 + 1) (x ^ 4-2x)
Get k (x) = 2x ^ 7 - 3x ^ 4 - 2x (^ is exponential sign)
Then K '(x) = 14x ^ 6 - 12x ^ 3-2
It's just a matter of directly deriving (differentiating) each function in it

1 / 1 + X / 6 power of E + X / 3 power of E + X / 2 power of e

y=1/[1+e^(x/6)+e^(x/3)+e^(x/2)]
dy=(-1)* [(1/6)e^(x/6)+(1/3)e^(1/3)+(1/2)e^(x/2)]*(1 / [1+e^(x/6)+e^(x/3)+e^(x/2)]^2) *dx

Let z = Z (x, y) be an implicit function determined by the equation xy = e ^ z-z

Let f (x, y, z) = xy-e ^ Z + Z
Fx=y
Fy=x
Fz=-e^z+1
therefore
az/ax=-Fx/Fz=-y/(-e^z+1)=y/(e^z-1)
az/ay=-Fy/Fz=-x/(-e^z+1)=x/(e^z-1)

Let y = y (x) be determined by the equation XY + e ^ y = 1, and find y "(0)

xy+e^y=1
e^y(0) =1
y(0) = 0
xy'+y+e^y y'=0
0+y(0) + y'(0) =0
y'(0) = 0
xy''+y'+ y' + e^y y'' + (y')^2e^y =0
0 +2y'(0)+ y''(0) + (y'(0))^2e^0 =0
y''(0) =0