- #1

IonizingJai

- 14

- 0

## Homework Statement

If ##x\sqrt{1+y} + y\sqrt{1+x } = 0##, then prove that ##\frac {dy} {dx} = \frac {-1}{(x-1)^2}##.

**2.Relevant Equations:**

$$ \frac {dy} {dx} =

- \frac {\left (\frac {\partial f}{\partial x} \right)} {\left( \frac {\partial f} {\partial y} \right)}.$$

3. The Attempt at a Solution :

$$ \frac {dy} {dx} =

- \frac {\left (\frac {\partial f}{\partial x} \right)} {\left( \frac {\partial f} {\partial y} \right)}.$$

3. The Attempt at a Solution :

Given expression : ##x\sqrt{1+y} + y\sqrt{1+x } = 0##, let's call it ##f(x)##.

Now, finding partial derivatives of ##f(x)## w.r.t x and y.

$$\frac {\partial f} {\partial x} = \sqrt{1+y} + \frac y {2\sqrt{1+x}}.$$

$$\frac {\partial f} {\partial y} = \frac x {2\sqrt{1+y}} + \sqrt{1+x}.$$

Now,

$$ \frac {dy} {dx} =

- \frac {\left (\frac {\partial f}{\partial x} \right)} {\left( \frac {\partial f} {\partial y} \right)}.$$

$$= - \frac {\left( \frac { 2\sqrt{1+x}\sqrt{1+y}~+y }{ 2\sqrt{1+x} } \right) } {\left( \frac { 2\sqrt{1+x}\sqrt{1+y}+~x }{ 2\sqrt{1+y} } \right)}$$

I'm stuck here. Since I don't know what should I do to get to the required result.