U = f (UX, V + y), v = g (U-V, V & # 178; y), F, g have first order continuous partial derivatives, find δ U / δ x, δ V / δ X. find the detailed explanation

For partial derivatives, other parameters can be regarded as constants
δu/δx
=f1' *δ(ux)/δx + f2' *δ(v+y)/δx
=f1' * x*δu/δx +f1' *u + f2' *δv/δx
and
δv/δx
=g1' *δ(u-v)/δx +g2' *δ(v²y)/δx
=g1' *(δu/δx- δv/δx) + 2vy *g2' *δv/δx
So we get the equations
(1- x *f1')δu/δx - f2' *δv/δx=f1' *u
g1' *δu/δx -(1+g1'-2vy*g2') *δv/δx=0
Then, by solving the system of quadratic equations of one variable,
δu/δx
= - f1' *u * (1+g1'-2vy*g2') / [f2' *g1' -(1- x *f1') *(1+g1'-2vy*g2')]
δv/δx
= - f1' *u *g1' / [f2' *g1' -(1- x *f1') *(1+g1'-2vy*g2')]

Z = f (XY, x + 2Y), find the second partial derivative, where f (U, V) is differentiable, the answer is y ^ 2F "11 + 2yf" 12
+I want to know how 2yf'12 is obtained,

f12=f21

Find the second derivative of XY sin (π y ^ 2) = 0 at (0, - 1)

y''=-(y'(x+2πycos(πy^2))-y(1+2πy'cos(πy^2)-2πysin(πy^2)*2πyy'))/(x+2πycos(πy^2))^2
=-(1/(2π)*(-2π)-(1-2π*1/(2π)-0))/(-2π)^2
=-(-1)/((2π)^2)
=1/4π^2

Given y = sin (XY), find its derivative

Solution:
From the derivation of the composite function, we can get the following results
Y'=COS(XY)*(XY)'
=COS(XY)*(Y+Y'X)
By substituting y = sin (XY) into the above formula, we can get:
Y'=(sin(xy)*cos(xy))/(1-x*cos(xy)) .

Finding the second derivative of XY sin (π y) = 0 at (0,1)

Derivation: y + XY '= π cos (π y) * y' further derivation: y '+ y' + XY '' = - π ^ 2Sin (π y) * (y ') ^ 2 + π cos (π y) * y' 'into (0,1) point from the first formula: 1 = - π y', y '= - 1 / π from the second formula: 2 * (- 1 / π) = - π * y' '= > y' '= 1 / π ^ 2

The second derivative of XY + sin (π y ^ 2) = 0 in (0,1)

Y + XY '+ cos (π y ^ 2) * 2 π y * y' = 0y '= - Y / (x + 2 π ycos (π y ^ 2)) y' (0,1) = - 1 / 2 π cos π = 1 / (2 π) y '' = - (y '(x + 2 π ycos (π y ^ 2)) - Y (1 + 2 π cos (π y ^ 2) - 2 π ysin (π y ^ 2) * 2 π YY') / (x + 2 π ycos (π y ^ 2)) ^ 2 =

Derivative y'x of implicit function y ^ 3-x ^ 2Y = 2

Derivations are made on both sides
3y²y'-2xy-x²y'=0
3y²y'-x²y'=2xy
y'=2xy/(2xy-x²)

Y ^ 3-x ^ 2Y = 2 for second derivative

The two-sided X-ray derivation: 3Y \35\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\178

U = x (Z + y) z = sin (x + Z) finding the second order partial derivative σ 2U / σ x σ y

Z is a function of X, so u is a function of X and y
z=sin(x+z) => z ' = cos(x+z) ( 1+ z ' ) => dz/dx = cos(x+z) / [ 1- cos(x+z) ]
u= F(x,y,z) = x(z+y),
δu/δx = δF/δx + δF/δz * dz/dx = z+y + x * cos(x+z) / [ 1- cos(x+z) ]
δ²u/δxδy = δ (δu/δx) /δy = 1

U = f (UX, V + y), v = g (U-V, V & # 178; y), F, g have first order continuous partial derivatives, find δ U / δ x, δ V / δ X. find the detailed explanation