Derivation of implicit function xy = e ^ (x + y) xy=e^(x-y) y+xy'=e^(x-y) *(1-y') y+xy'=xy-xy*y' (x+xy)y'=xy-y y'=(xy-y)/(x+xy) My question is, how can the second step be obtained from the first step?

Derivation of implicit function xy = e ^ (x + y) xy=e^(x-y) y+xy'=e^(x-y) *(1-y') y+xy'=xy-xy*y' (x+xy)y'=xy-y y'=(xy-y)/(x+xy) My question is, how can the second step be obtained from the first step?

According to the title, what the owner needs is the second step
xy=e^(x-y)
y+xy'=e^(x-y) *(1-y')
Derivative x (take y as a constant) and then y (take x as a constant)
The results are as follows:
Left: y + XY '
On the right: e ^ (X-Y) * (1-y '), this step is to take X-Y as a whole; first, the derivative of X: e ^ (X-Y) and then the compound derivative is: 1;
Then the derivative of Y is e ^ (X-Y) * - y ';
The results show that: e ^ (X-Y) (1-y ')

Derivation of sinxy + ln (y + x) = x

cos(xy) ×（y+xy'）+1/(x+y) ×（1+y'）=1
(xcosxy +1/(x+y))y'=1-ycosxy-1/(x+y)
y'=[1-ycosxy-1/(x+y)]/(xcosxy +1/(x+y))

Find the derivative of the implicit function e ^ y + xy-e = 0,

The function determined by the equation E ^ y + xy-e = 0 is y = f (x),
Therefore, when deriving x from both sides of the equation, we should treat y as a function of X, so that we can get the result
e^y*y'+y+xy'=0
So y '= - Y / (e ^ y + x)
Note: y '= dy / DX

The implicit function can take logarithm first and then be derivative
xy=e^(x+y)
If you don't take the logarithm first,
(e^(x+y)-y)/(x-e^(x+y))

xy=e^(x+y)
y+xy′=e^(x+y)（1+y′）
y′=（y-e^(x+y)）/（-e^(x+y）-x）

How to find the derivative of xy = e ^ (x, y) implicit function

Constructor, f (x, y) = xy-e ^ (XY)
Then dy / DX = - FX / FY = - [y-e (XY) * y] / [x-e ^ (XY) * x]

How to find the derivative of the inverse function of y = x · e ^ x

y'=dy/dx=(x+1)·e^x
The derivative of its inverse function is
dx/dy=1/y'=1/[(x+1)·e^x]

(arctanx) derivative = 1 / x2 + 1
How to seek derivation

Let x = tant, then t = arctanx, DX = [(COS & # 178; t + Sin & # 178; t) / (COS & # 178; x)] DT DX = (1 / cos & # 178; t) DT / DX = cos & # 178; t DT / DX = 1 / (1 + Tan & # 178; t) because x = tant, the above formula t '= 1 / (1 + X & # 178;)

Y = x times the derivative of y plus the square of the derivative of Y

y=(xy)'+y'^2=y+xy'+y'^2
xy'+y'^2=0
x+y'=0
y'=-x
y=(-x^2)/2 + c

How to find the derivative of 4 times the square of x minus x times y plus the square of Y minus 6
I hope you can answer for me,

The derivative of X is 8x-y and the derivative of Y is - x-2y

The maximum value of the function y = x-sinx, X ∈ [π 2, π] is______ ．

∵ y = x monotonically increases on [π 2, π], y = - SiNx monotonically increases on [π 2, π]; y = x-sinx monotonically increases on [π 2, π], that is, the maximum value is f (π) = π, so the answer is π