Math, asked by pallavishendge8125, 1 year ago

if u=log (x3+y3+z3-3xyz)show that (d/dx+d/dy+d/z)2u = -9/(x+y+z)2

Answers

Answered by himonika2003
13

Answer:

Step-by-step explanation:

Attachments:
Answered by lublana
50

Given:

u=log(x^3+y^3+z^3-3xyz)

To prove that

(d/dx+d/dy+d/dz)2u=-\frac{9}{(x+y+z)^2}

Solution:

u=log(x^3+y^3+z^3-3xyz)

\frac{du}{dx}=\frac{3x^2-3yz}{x^3+y^3+z^3-3xyz}

\frac{du}{dy}=\frac{3y^2-3xz}{x^3+y^3+z^3-3xyz}

\frac{du}{dz}=\frac{3z^2-3xy}{x^3+y^3+z^3-3xyz}

(d/dx+d/dy+d/dz)u=\frac{d(u)}{dx}+\frac{d(u)}{dy}+\frac{d(u)}{dz}

(d/dx+d/dy+d/dz)(u)=\frac{(3x^2-3yz)}{x^3+y^3+z^3-3xyz}+\frac{(3y^2-3xz)}{x^3+y^3+z^3-3xyz}+\frac{(3z^2-3xy)}{x^3+y^3+z^3-3xyz}

(d/dx+d/dy+d/dz)u=\frac{3x^2-3yz+3y^2-3xz+3z^2-3xy}{x^3+y^3+z^3-3xyz}

(d/dx+d/dy+d/dz)u=(3\frac{x^2+y^2+z^2-xy-yz-xz}{x^3+y^3+z^3-3xyz})

(d/dx+d/dy+d/dz)u=\frac{3(x^2+y^2+z^2-xy-yz-xz}{(x+y+z)(x^2+y^2+z^2-xy-yz-xz)}

(d/dx+d/dy+d/dz)u=\frac{3}{x+y+z}

(d/dx+d/dy+d/dz)^2u=-\frac{3}{(x+y+z)^2}-\frac{3}{(x+y+z)^2}-\frac{3}{(x+y+z)^2}

(d/dx+d/dy+d/dz)^2u=-\frac{9}{(x+y+z)^2}

Hence, proved.

Similar questions