Y = Xe ^ y, find dy / DX | x = 0

Y = Xe ^ y, find dy / DX | x = 0

y'=(xe^y)'
=x'e^y +x(e^y)'
=e^y+xe^y y'
y‘=e^y/(1-e^y)
There seems to be no definite value for dy / DX = e ^ y / (1-e ^ y) x = 0

Let y = y (x) be the implicit function determined by the functional equation ln (x ^ 2 + y ^ 2) = x + Y-1, and find dy / DX

ln(x ²+ y ²)= x+y-1
Derivative of X on both sides:
(2x+2yy ')/(x ²+ y ²)= 1+y '
Sorting:
y '=(2x-x ²- y ²)/ (x ²+ y ²- 2y)
so
dy/dx=(2x-x ²- y ²)/ (x ²+ y ²- 2y)

Let the function y be determined by the equation ln y + X / y = 0, and find dy / DX

ln y+x/y=0
Derivation on both sides of the equation:
y'*1/y+1/y+x*y'(-1/y ²)= 0
(1/y-x/y ²) y'=-1/y
∴y'=(-1/y)/(1/y-x/y ²)=- y/(y-x)
∴dy/dx=-y/(y-x)

Given the function y = ln [x + (1 + x ^ 2) ^ (1 / 2)], then DX / dy

y=ln[x+(1+x^2)^(1/2)],
dy/dx={1/[x+(1+x^2)^(1/2)]}*【x+(1+x^2)^(1/2)】‘
={1/[x+(1+x^2)^(1/2)]}*【1+x/(1+x^2)^(1/2)】
=1/(1+x^2)^(1/2)
therefore
dx/dy=(1+x^2)^(1/2)

Let y = (tan2x) ^ cot (x / 2), find dy / DX

There seems to be a mistake upstairs. Be careful
Take logarithms on both sides and get
lny=ln【(tan2x)^cot(x/2) 】=cot(x/2)ln（tan2x）
Then take the derivative on both sides, and you get
y'/y={-[csc（x/2)]^2*ln(tan2x)}/2+{2cot(x/2)(sec2x)^2}/tan2x
Namely
y'=y{-[csc（x/2)]^2*ln(tan2x)}/2+{2cot(x/2)(sec2x)^2}/tan2x
yes
y'=dy/dx =(tan2x)^cot(x/2)*{-[csc（x/2)]^2*ln(tan2x)}/2+{2cot(x/2)*(sec2x)^2}/tan2x

X = sin (Y / x) + e ^ 2 find dy / DX

X = sin (Y / x) + e ^ 2 find dy / DXD (x) = D (sin (Y / x) + e ^ 2) DX = DSIn (Y / x) + de ^ 2DX = cos (Y / x) d (Y / x) DX = cos (Y / x) (XDY YDX) / x ^ 2x ^ 2DX = xcos (Y / x) dy ycos (Y / x) DXX ^ 2DX + ycos (Y / x) DX = xcos (Y / x) dydy / DX = (x ^ 2 + ycos (Y / x)) / xcos (Y / x) sin (Y / x) = x -