# Implicit differentiation problem

## 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 : $x\sqrt{1+y} + y\sqrt{1+x } = 0$, lets 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.

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kuruman
Homework Helper
Gold Member
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
Mentor

## 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.
IonizingJai said:
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 : $x\sqrt{1+y} + y\sqrt{1+x } = 0$, lets 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.

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

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
Homework Helper
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 $\frac {dy} {dx} = \frac {-1}{(x-1)^2}$

$\frac{dy}{dx} = -\frac{1}{(1+x)^2}$, 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
Homework Helper
EDIT : the required result was $\frac{dy}{dx} = -\frac{1}{(1+x)^2}$.
All is clear now

Mark44
Mentor
$$\frac {x} {y} = -\frac{\sqrt{1+x}}{\sqrt{1+y}}$$
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
Homework Helper
$$y = -\frac{x}{1+x}$$ .
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?

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
Homework Helper
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?

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