Implicit differentiation problem

[IonizingJai]

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 :

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.

 

[kuruman]

You have not used the given fact that $f(x,y)=0$ so far. I am sure there is some clever substitution that will allow you to eliminate the $y$ variable.

 

[Mark44]

Homework Statement

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

Your thread title says that this is an implicit differentiation problem. Why don't you do it that way?
Just take the derivative with respect to x of each term on the left side of your equation. There is no need for partial derivatives, as the assumption in implicit differentiation is that, in this case, y is a differentiable function of x.

Also, this is a purely math problem, so I am moving it to the calculus section.

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 :

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.

 

[Porkaborg]

Are you sure that this is true ?

dydx=−(∂f∂x)(∂f∂y)

dy/dx tells us to derive the function y=f(x) with respect to x.

 

[Mark44]

Are you sure that this is true ?

dy/dx tells us to derive the function y=f(x) with respect to x.

Yes, the equation as stated in post #1 is correct.

If $f(x, y) = 0$, the the total differentials of both sides have to be equal.
$\frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy = d(0) = 0$
$\Rightarrow \frac{\partial f}{\partial y}dy = -\frac{\partial f}{\partial x}dx$
$\Rightarrow \frac{dy}{dx} = -\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$

IMO, though, partials are overkill in this problem, but using ordinary implicit differentiation still ends up with a messy answer that doesn't look like what is supposed to be proved.

 

[kuruman]

You can move one radical in the given expression to the other side, square, solve the quadratic for $y$ and throw out the solution that does not satisfy the given expression. Then just take the derivative of the good solution as @Mark44 suggested. I don't agree that it's messy, but I agree that you don't get the answer that you are asked to show.

 

[Mark44]

I don't agree that it's messy, but I agree that you don't get the answer that you are asked to show.

That's pretty much what I meant by calling it "messy."

 

[epenguin]

I would do what kuruman suggests only if I had to. As you are invited to get rid of y, and as what causes most of he trouble in this sort of calculation are √ terms, I would start by applying your first equation in 1.
$x\sqrt{1+y} + y\sqrt{1+x } = 0$
not even to your equation for dy/dx but better to

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

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

to get rid of √(1 +y) and see where that leads you.

 

[epenguin]

Or another idea. Get rid of square root terms in the original equation 1 and then do the implicit differentiation.

 

[IonizingJai]

You have not used the given fact that $f(x,y)=0$ so far. I am sure there is some clever substitution that will allow you to eliminate the $y$ variable.

Thanks to all who replied.

Here's what I did, as kuruman suggested,
First, covert given expression to
$$\frac {x} {y} = -\frac{\sqrt{1+x}}{\sqrt{1+y}} $$
Then, squaring b/s we get, $\frac {x^2} {y^2} = -\frac{1+x}{1+y}$, manupulating expression to get , $x+ y +xy = 0$.
$$ y = -\frac{x}{1+x}$$ .
Applying condition $\frac{1-x}{1-y} \gt 0$ we get $x\gt -1$.
(I have further doubt here, wherther to apply condition for $\frac {x} {y} = -\frac{\sqrt{1+x}}{\sqrt{1+y}}$ or substitute for y in x in the given expression $x\sqrt{1+y} + y\sqrt{1+x } = 0$ in order to throw out the intervals in which the parent equation becomes invalid. )
Anyway, we diffrentiate the equation $y = -\frac{x}{1+x}$ to get $\frac{dy}{dx}$ and the result as required.

 

[ehild]

Anyway, we diffrentiate the equation $y = -\frac{x}{1+x}$ to get $\frac{dy}{dx}$ and the result as required.

$\frac{dy}{dx} = -\frac{1}{(1+x)^2}$, and the problem said

prove that $\frac {dy} {dx} = \frac {-1}{(x-1)^2}$

 

[IonizingJai]

$\frac{dy}{dx} = -\frac{1}{(1+x)^2}$, and the problem said

Sorry, I made the mistake in OP and it went unnoticed even after I gave it a check, I find writing LaTeX code difficult, apologies for any time waste, my mistake might have caused.

I will edit the OP, okay?
Looks like there is no option to edit the post.

so, I will post it here:
EDIT : the required result was $\frac{dy}{dx} = -\frac{1}{(1+x)^2}$.

 

[ehild]

EDIT : the required result was $\frac{dy}{dx} = -\frac{1}{(1+x)^2}$.

All is clear now

 

[Mark44]

$$\frac {x} {y} = -\frac{\sqrt{1+x}}{\sqrt{1+y}} $$
Then, squaring b/s we get, $\frac {x^2} {y^2} = -\frac{1+x}{1+y}$

If you square $-\frac{\sqrt{1+x}}{\sqrt{1+y}}$, you get $\frac{1+x}{1+y}$, not $-\frac{1+x}{1+y}$.

 

[ehild]

$$ y = -\frac{x}{1+x}$$ .
Applying condition $\frac{1-x}{1-y} \gt 0$ we get $x\gt -1$.
(I have further doubt here, wherther to apply condition for $\frac {x} {y} = -\frac{\sqrt{1+x}}{\sqrt{1+y}}$ or substitute for y in x in the given expression $x\sqrt{1+y} + y\sqrt{1+x } = 0$ in order to throw out the intervals in which the parent equation becomes invalid. )

x>-1 and y-1 should hold otherwise you get zero or negative number under the square roots.
y=-x/(1+x) = -1+1/(x+1) . Is it true that y>-1 whenever x>-1?

 

[IonizingJai]

Oh no, I just realized I have made so many mistakes while writing out the reply.

If you square $-\frac{\sqrt{1+x}}{\sqrt{1+y}}$, you get $\frac{1+x}{1+y}$, not $-\frac{1+x}{1+y}$.

Correct.

EDIT: the condition is $\frac {1+x} {1+y} \gt 0$.

Still, not fully clear on inequality/condition part.

 

[ehild]

EDIT: the condition is $\frac {1+x} {1+y} \gt 0$.

Still, not fully clear on inequality/condition part.

What condition ? (1+x)/(1+y) is always greater than zero if y=-x/(1+x) . Is it possible that x=-10, for example?
The starting equation is valid only when the expressions under the square roots are not negative, that is, both x>-1 and y>-1 must hold.
You can choose x>-1, prove that then y>-1.
Your solution can be rewritten as
y=-x/(1+x) = -1+1/(x+1) . Is it true that y>-1 whenever x>-1?