Question

if u= tan inverse (xy/square root 1+x²+y²) then show that del²u/del x del y = 1/(1+x²+y²) to the power 3/2​

Answer 1

Answer:

Given, u=tan

−1

(

x+y

x

2

+y

2

)

⇒tanu=

(x+y)

x

2

+y

2

On differentiating both the sides w.r.t. x and y respectively, we get

sec

2

u

dx

du

=

(x+y)

2

2x(x+y)−(x

2

+y

2

)

....(1)

and sec

2

u

dy

du

=

(x+y)

2

2y(x+y)−(x

2

+y

2

)

....(2)

Multiplying x with equation (1) and y to equation (2).

Therefore, xsec

2

u

dx

du

=

(x+y)

2

x

3

+2x

2

y−xy

2

....(3)

and ysec

2

u

dy

du

=

(x+y)

2

y

3

+2xy

2

−x

2

y

....(4)

Adding equations (3) and (4), we get

sec

2

u(x

dx

du

+y

dy

du

)=

(x+y)

2

x

3

+2x

2

y−xy

2

+y

3

+2xy

2

−x

2

y

=

(x+y)

2

x

3

+x

2

y+y

3

+xy

2

=

(x+y)

2

(x+y)(x

2

+y

2

)

=

x+y

x

2

+y

2

=tanu

Thus x

dx

du

+y

dy

du

=sinucosu

=

2

1

sin2u

Answer 2

Given:

• u = $tan^{-1} (xy/\sqrt{1+x^{2} +y^{2} } )$

To Find:

• Show that  $\frac{del^{2} (u)}{del(x)del(y)} = \frac{1}{(1+x^{2} +y^{2})^3/2 }$

Solution:

• first, differentiate $\frac{delu }{dely}$ with respect to y.
• we are partially differentiating u with respect to y and we get the following equation.
• $\frac{delu }{dely} = \frac{1}{1+\frac{x^{2}+ y^{2} }{1+x^{2} +y^{2} } } *x*\frac{(\sqrt{1+x^{2} +y^{2} } - y*\frac{1}{2}*(1+x^{2} +y^{2})^{-1/2} *2y }{1+x^{2} +y^{2} }$
• On further simplification we get,
• $\frac{delu }{dely} = \frac{x}{1+x^{2} +y^{2}+x^{2} +y^{2} } *(\frac{1+x^{2} +y^{2} - y^{2} }{\sqrt{1+x^{2} +y^{2} } } )$  
• Taking LCM and simplifying we get,
• $\frac{delu }{dely} = \frac{x}{(1+y^{2}) *\sqrt{1+x^{2} +y^{2} } }$
• In the next step, we are differentiating the above obtained equation as we need to prove.
• $\frac{del^{2}u }{delxdely} = \frac{del(\frac{x}{(1+y^{2}) *\sqrt{1+x^{2} +y^{2}) } } }{delx}$
• In the above step, we are double partially differentiating with respect to x.
• $\frac{del^{2}u }{delxdely} = (\frac{1}{1+y^{2} })*(\frac{(1+x^{2} +y^{2}) -x^{2} }{(1+x^{2} +y^{2} )^{3/2} } )$

• $\frac{del^{2}u }{delxdely} = \frac{1}{(1+x^{2}+y^{2})^{3/2} }$

Hence Proved.

$\frac{del^{2}u }{delxdely} = \frac{1}{(1+x^{2}+y^{2})^{3/2} }$