Question

if u=(x²+y²+z²)^-1/2, find the value of (d²u/dx²+d²u/dy²+d²u/dz²)?​

Answer 1

Answer:

u=

x

2

+y

2

+z

2

dx

du

=

2

x

2

+y

2

+z

2

1

(2x)=

x

2

+y

2

+z

2

x

dx

2

d

2

u

=

(

x

2

+y

2

+z

2

)

2

x

2

+y

2

+z

2

dx

d

(x)−x

dx

d

(

x

2

+y

2

+z

2

)

=

x

2

+y

2

+z

2

x

2

+y

2

+z

2

−x

2

x

2

+y

2

+z

2

1

(2x)

=

(x

2

+y

2

+z

2

)

x

2

+y

2

+z

2

x

2

+y

2

+z

2

−x

2

=

(x

2

+y

2

+z

2

)

3/2

y

2

+z

2

Similarly,

dy

2

d

2

u

=

(x

2

+y

2

+z

2

)

3/2

x

2

+z

2

,

dz

2

d

2

u

=

(x

2

+y

2

+z

2

)

3/2

x

2

+y

2

u(

dx

2

d

2

u

+

dy

2

d

2

u

+

dz

2

d

2

u

)=

x

2

+y

2

+z

2

(

(x

2

+y

2

+z

2

)

3/2

2(x

2

+y

2

+z

2

)

).

=

(x

2

+y

2

+z

2

)

3/2

(x

2

+y

2

+z

2

)

3/2

.2

=2.

solution