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Find that the partial derivative z = f (XY, x ^ 2 + y ^ 2) has a second-order continuous partial derivative Find a ^ 2Z / ax ^ 2 A ^ 2Z / axay
Dz/Dx = f1*y + f2*(2x) = y*f1 + 2x*f2,D ² z/Dx ² = (D/Dx)(y*f1 + 2x*f2) = [y*(y*f11 + 2x*f12)+ 2*f2 + 2x*(y*f21 +2x*f22)] = ……,D ² z/DxDy = (D/Dy)(y*f1 +2x*f2) = [f1 + y*(x*f11 + 2y*f12)+ ...
X = (y, z), y = (x, z), z = Z (x, y) is a function with continuous partial derivative determined by F (x, y, z) = 0. It is proved that x to y partial derivative * y to Z partial derivative * Z to x partial derivative = - 1
When x = f (y, z)
δ F/ δ y=F'1* δ x/ δ y+F'2=0
Namely: δ x/ δ y=-F'2/F'1
Similarly: δ y/ δ z=-F'3/F'2, δ z/ δ x=-F'1/F'3
Therefore（ δ x/ δ y)*( δ y/ δ z)*( δ z/ δ x)
=(-F'2/F'1)*(-F'3/F'2)*(-F'1/F'3)=-1
The second derivative is equal to 0. The tangent passes through the function to prove that the tangent passes through the function at a tangent point of a function
Y = x ^ 3 the tangent at x = 0 passes through the function,
College senior two, on the derivative formula of binary implicit function: why is there this formula! y‘=-f'（x）/f'（y） Why is there this formula? How did you deduce it?
Binary implicit function f (x, y) = 0
Fully differential the above formula: FX (x, y) DX + FY (x, y) dy = 0 (where FX (x, y) means that f (x, y) calculates the partial differential of x)
So dy / DX = - FX (x, y) / FY (x, y)
In your way, y '= - f' (x) / F '(y) correspond to the above formula respectively
..
Find the derivative of the implicit function determined by the implicit function y-x-1 / 2siny = 0 in the higher number at x = 0 When x = 0 and y = 0 are substituted into the equation y'-1-1/2(cosy)y'=0 -Where did you come from thanks
Both sides of the equation derive x at the same time. When you encounter y, you should know that y is a function of X. use the chain rule to get y '- 1-1 / 2 (cosy) y' = 0, and then substitute x = 0 and the corresponding y = 0 to get y '(0) - 1-0.5 × one × Y '(0) = 0, so y' (0) = 2
Finding the partial derivative of implicit function Siny + e ^ x-xy ^ 2 = 0, find dy / DX
solution
Two side derivation
y‘cosy+e^x-y^2-2xyy'=0
Namely
y’(cosy-2xy)=y^2-e^x
y'=(y^2-e^x)/(cosy-2xy)
perhaps
F(x,y)=siny+e^x-xy^2=0
Fx=e^x-y^2
Fy=cosy-2xy
dy/dx=-Fx/Fy=(y^2-e^x)/(cosy-2xy)
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