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Implicit differentiation problem

  • #1

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.
 

Answers and Replies

  • #2
kuruman
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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.
 
  • #3
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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.
 
  • #4
3
0
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.
 
  • #5
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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.
 
  • #6
kuruman
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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.
 
  • #7
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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."
 
  • #8
epenguin
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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.
 
  • #9
epenguin
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Or another idea. Get rid of square root terms in the original equation 1 and then do the implicit differentiation.
 
  • #10
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.
 
  • #11
ehild
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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}##
 
  • #12
##\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}##.
 
  • #13
ehild
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EDIT : the required result was ##\frac{dy}{dx} = -\frac{1}{(1+x)^2}##.
All is clear now :smile:
 
  • #14
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$$\frac {x} {y} = -\frac{\sqrt{1+x}}{\sqrt{1+y}} $$
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}##.
 
  • #15
ehild
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$$ y = -\frac{x}{1+x}$$ .
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?
 
  • #16
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.
 
  • #17
ehild
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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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