Finding the partial derivative of Z to Y: z = (1 + XY) ^ y

Finding the partial derivative of Z to Y: z = (1 + XY) ^ y

z=(1+xy)^y=e^[y×ln(1+xy)]
So the partial derivative of Z to y
?z/?y=e^[y×ln(1+xy)]×?[y×ln(1+xy)]/?y
=?[y×ln(1+xy)]/?y×(1+xy)^y
=[(1+xy)^y]×ln(1+xy)+y×(1+xy)^y×?[ln(1+xy)]/?y
And? [ln (1 + XY)] /? Y = x / (1 + XY)
So? Z /? Y = [(1 + XY) ^ y] × ln (1 + XY) + y × [(1 + XY) ^ y] × X / (1 + XY)
=[(1+xy)^y]×ln(1+xy)+xy×(1+xy)^(y-1)
Multiplier sign and X are a bit similar. Pay attention to the difference

The partial derivative of Z = arctan [(x + y) / (1-xy)]

How to find the partial derivative of Z = (1 + XY) ^ x for y?
As the title shows
Wrong, the topic should be z = (1 + XY) ^ y, the partial derivative of Y

In the derivation of X, y is regarded as a constant: Y (1 + XY) ^ (Y-1)
For the derivation of Y, take x as a constant, use the method of natural logarithm on both sides of the equation, or use the equivalent deformation of y = e ^ LNX for the derivation: (1 + XY) ^ y. (LN (1 + XY) + (y.x) / (1 + XY))

Calculating two first order partial derivatives of Z = y (e ^ - XY)

Partial derivative of X: ZX = Ye ^ (- XY) * (- y) = (- y ^ 2) (e ^ - XY)
The partial derivative of Y: ZY = e ^ (- XY) + Ye ^ (- XY) * (- x) = e ^ (- XY) - XYE ^ (- XY)
Explanation:
To find the partial derivative of X is to treat x as a variable and y as a constant
To find the partial derivative of Y is to treat y as a variable and X as a constant

Z = (1 + XY) ^ y, find the partial derivative of Y

If we take the natural logarithm of both sides at the same time, we have
lnZ=yln(1+xy)
The partial derivatives of the two sides are obtained simultaneously, and the results are simplified
Zy=(ln(1+xy)+xy/(1+xy))*(1+xy)^y

For the second partial derivatives zxx, zyy and ZXY, z = arctan [(x + y) / (1-xy)],

The answers are as follows:

The first partial derivative of Z = ln (TaNx / y)
Finding the total differential of Z = arctanx + Y / X-Y

(1) z=ln(tanx/y)
dz/dx=1/(tanx/y)*(sec²x/y)=sec²x/tanx
dz/dy=1/(tanx/y)*(-tanx/y²)=-1/y
(2) z=arctanx+y/x-y
dz=(dz/dx)*dx+(dz/dy)*dy
=[1/(x²+1)-y/x²]dx+(1/x-1)dy

Seeking partial derivative in Higher Mathematics
Let z = Z (x, y) be determined by the equation LNX + XYZ + lnz = 0, and find the partial Z / partial y

LNX + XYZ + lnz = 0 the partial derivative of Y on both sides of equal sign
Three terms on the left side of the equal sign seek the derivative of Y (take x as a constant)
First item: 0
The second term: X (Z + y * partial Z / partial y)
Third: 1 / Z * partial Z / partial y
The sum of three terms equals 0
figure out
Partial Z / partial y = - XZ ^ 2 / (1 + XYZ)

Some simple problems of partial derivative
A total of seven, the best step, the best text,

The steps are as follows,
 

The second derivative of y = ln (1 + x ^ 2),

y' = (1+x²)'/(1+x²) = 2x/(1+x²)
y" = [(1+x²)(2x)' - (1+x²)'(2x)/(1+x²)²
= 2(1-x²)/(1+x²)²