Implicit differentiation problem

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

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.
 
Physics news on Phys.org
  • #2
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
IonizingJai said:

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##, 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.
 
  • Like
Likes scottdave
  • #4
Are you sure that this is true ?
IonizingJai said:
dydx=−(∂f∂x)(∂f∂y)
dy/dx tells us to derive the function y=f(x) with respect to x.
 
  • #5
Porkaborg said:
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.
 
  • Like
Likes Porkaborg
  • #6
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
kuruman said:
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
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

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

$$\frac {\partial f} {\partial y} = \frac x {2\sqrt{1+y}} + \sqrt{1+x}.$$
to get rid of √(1 +y) and see where that leads you.
 
  • #9
Or another idea. Get rid of square root terms in the original equation 1 and then do the implicit differentiation.
 
  • #10
kuruman said:
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
IonizingJai said:
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
ehild said:
##\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
IonizingJai said:
EDIT : the required result was ##\frac{dy}{dx} = -\frac{1}{(1+x)^2}##.
All is clear now :smile:
 
  • #14
IonizingJai said:
$$\frac {x} {y} = -\frac{\sqrt{1+x}}{\sqrt{1+y}} $$
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
IonizingJai said:
$$ y = -\frac{x}{1+x}$$ .
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.
Mark44 said:
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
IonizingJai said:
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?
 
Last edited:

1. What is implicit differentiation?

Implicit differentiation is a method used in calculus to find the derivative of a function that is not explicitly written in terms of one variable. It allows us to find the rate of change of a dependent variable with respect to an independent variable.

2. When should implicit differentiation be used?

Implicit differentiation should be used when it is difficult or impossible to solve for one variable in a function that contains multiple variables. It is also useful when dealing with equations involving a mixture of trigonometric, exponential, and logarithmic functions.

3. How does implicit differentiation work?

To use implicit differentiation, we treat the dependent variable as a function of the independent variable and then apply the chain rule to differentiate both sides of the equation. This allows us to find the derivative of the dependent variable with respect to the independent variable.

4. What is the difference between implicit differentiation and explicit differentiation?

Explicit differentiation is used when the dependent variable is explicitly written in terms of the independent variable, while implicit differentiation is used when the dependent variable is not explicitly written in terms of the independent variable. In explicit differentiation, we can simply apply the power rule and other basic differentiation rules, while implicit differentiation requires the use of the chain rule.

5. Are there any limitations to implicit differentiation?

Yes, implicit differentiation can only be used to find the derivative of a function with respect to a single variable. It cannot be used to find higher order derivatives or partial derivatives. Additionally, it may not always be possible to find an explicit expression for the derivative using implicit differentiation, so numerical methods may be needed.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
681
  • Calculus and Beyond Homework Help
Replies
5
Views
754
  • Calculus and Beyond Homework Help
Replies
5
Views
612
  • Calculus and Beyond Homework Help
Replies
6
Views
746
  • Calculus and Beyond Homework Help
Replies
21
Views
826
  • Calculus and Beyond Homework Help
Replies
6
Views
839
  • Calculus and Beyond Homework Help
Replies
6
Views
543
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
760
  • Calculus and Beyond Homework Help
Replies
4
Views
555
Back
Top