If u = tan^(-1){(x^3 + y^3)/(x - y)}. Prove that?

Question

x*(partial derivative w.r.t. x) + y*(patial derivative w.r.t. y) = sin2u

Answer 1

Let z be the argument of the inverse tangens function
z = (x³+y³)/(x-y)
and
u = tan⁻¹(z) = arctan(z) <=> z = tan(u) = sin(u)/cos(u)

The left hand side takes the form:
x·∂u/∂x + y·∂u/∂y
= x·∂u/∂z·∂z/∂x + y·∂u/∂y·∂z/∂y
= x·1/(1+z²)·(3x²/(x-y) - (x³+y³)/(x-y)²) + y·1/(1+z²)·(3y²/(x-y) + (x³+y³)/(x-y)²)
= 1/(1+z²)·(3x³/(x-y) + 3y³/(x-y) - (x-y)·(x³+y³)/(x-y)²)
= 1/(1+z²)·2·(x³+y³)/(x-y)
= 2z/(1+z²)
(insert z = sin(u)/cos(u)
= 2·sin(u)/cos(u)/(1+ [sin(u)/cos(u)]²)
= 2·sin(u)·cos(u)/(cos²(u) + sin²(u))
(because of the pythagorean identity sin²(u)+cos²(u)=1)
= 2·sin(u)·cos(u)
(double angle formula sin(2u) = 2 sin(u)·cos(u)
= sin(2u) q.e.d

Answer 2

tan u = (x³+y³)/(x-y) 1st differentiating partially wrt x then y

sec^2 u * ∂u/∂x = (3x^2)(x-y)-(x³+y³) / (x-y)² .....(1)
sec^2 u * ∂u/∂y = (3y^2)(x-y) +1(x³+y³) / (x-y)² ..... (2)

x * (1) + y* (2) adding with multiplication

sec^2 u *[x ∂u/∂x+y ∂u/∂y]
=[2x^4 -2 y x^3 + 2 x y^3 - 2y^4]/(x-y)^2
= 2 (x-y) (x^3+y^3) / (x-y)^2
= 2 * tan u

[x ∂u/∂x+y ∂u/∂y] = 2 * tan u / sec^2 u = sin (2u) answer

Answer 3

to educate 2 sin y cos y = ( 2 tan y) / (a million + tan² y) RHS = ( 2 tan y) / (a million + tan² y) =(2siny/comfortable)/(a million +sin² y/cos² y) =(2siny/comfortable){cos²y/(cos² y +sin² y)} =(2sinycosy/(a million} =2sinycosy =LHS

Answer 4

It's messy (Boy! is it messy ☺) but just take the partials, do the indicated multiplications, and you can get it to collapse to 1/csc(2u) = sin(2u)


Doug

Answer 5

真正地