Let the function y = y (x) be determined by the equation y square - 2XY = 7, and find dy / DX Let the function y = y (x) be determined by the equation y ^ 2-2xy = 7, and find dy / DX. Thank you very much

Let the function y = y (x) be determined by the equation y square - 2XY = 7, and find dy / DX Let the function y = y (x) be determined by the equation y ^ 2-2xy = 7, and find dy / DX. Thank you very much

The differential of Y ^ 2-2xy = 7 is obtained
2ydy-2(ydx+xdy)=0,
∴（y-x)dy=ydx,
∴dy/dx=y/(y-x).

Let the function y = LNX square, and find dy|x of DX = 1?

Y = Linx ² Do you
dy/dx=1/x ². （x ²）'
=1/x ². (2x)
=2x/x ²
dy/dx|x=1=2*1/1 ²= two

Let the function y = f (x) be determined by x square + y fourth power + X + 2Y = 1, and find the dy of DX

d(x ²)+ d(y^4)+dx+d(2y)=d(1)
2xdx+4y ³ dy+dx+2dy=0
So dy / DX = - (2x + 1) / (4Y) ³+ 2)

The mathematical problem x = x (y) is the inverse function of y = y (x), dy / DX = Xe ^ x, when x > 0, the ball d ^ 2x / dy2

{if the inverse function of F (x) is g (x), then f '(x) = 1 \ G' (x)}
Dy \ DX = Xe ^ x, then the first derivative of the original function is 1 \ Xe ^ X
Then d ^ 2x \ dy ^ 2 = (1 \ Xe ^ 2) '= - (1 + x) \ (e ^ x * x ^ 2)

Let y = y (x) be determined by y-xe ^ y = 1, and find dy / DX

y-xe^y=1
y ' - [x ' e^y + x（e^y）'] = 0
y ' - [ e^y + x y ' e^y ] = 0
（1 - x e^y）y ' = e^y
y ' = e^y /（1 - x e^y）

The function y = y (x) is determined by the equation y = 1-xe ^ y, and find dy / DX x = 0

Taking the derivative of X on both sides (note that y here is a function of x):
y'=-e^y-(xe^y)*y'；
Sorting:
y'=-e^y/(1+xe^y)
It can be seen from the original formula that when x = 0, y = 1, brought into the above formula, y '= - E
-E is the answer