# Can I Differentiate Both Sides to Solve y in y^2=ln(1-y^2)?

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• NODARman
In summary: So the only extremum is at ##x=y=0##.In summary, the conversation discusses finding extreme points using the "z" function and solving for "y" in the system using "ln" (natural logarithm). It is suggested to use polar coordinates and solve for the value of "r" at which "z" is maximal, leading to a solution of "r=0" and an extremum at "x=y=0". It is also noted that equations of the form (polynomial) = (exponential) must be solved numerically. The problem at hand, y^2=ln(1-y^2), can be solved by setting both sides equal to 0, resulting in the only extrem
NODARman
TL;DR Summary
How can I find the extrema of the function?
Hi, I'm differentiating the "z" function to find extreme points but after solving the first partial derivatives with respect to "x" and "y" and also the "x" variable of the system, I can't find "y" (still in the system) using "ln" (natural logarithm).

The question is can I differentiate both sides and easily write what "y" is or I can't use that method to solve it?

The problem: y^2=ln(1-y^2)

This is how I've got here (picture):

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You have $z = (x^2 + y^2)e^{-(x^2 + y^2)} - (x^2 + y^2)$. It would make sense here to use polar coordinates instead, and find the value of $r$ at which $z$ is maximal. This leads to $$1 - r^2 = e^{r^2}$$ which is what you obtained for $y$ after concluding that $x = 0$.

Now by inspection $r = 0$ is a solution, and since the left hand side is less than 1 for $r > 0$ and the right hand side is greater than 1 for $r > 0$ the only solution is $r = 0$. Thus the extremum is at $x = y = 0$.

That one could be solved analytically, but in general equations of the form $$\mbox{(polynomial)} = \mbox{(exponential)}$$ have to be solved numerically.

Last edited:
NODARman said:
The problem: y^2=ln(1-y^2)
You can solve this particular equation similarly to the way @pasmith did above for ##r##. The lefthand side is non-negative while the righthand side is non-positive. The only possible solution is when both sides equal 0, which is at ##y=0##.

WWGD

## 1. How do I solve for y in y^2=ln(1-y^2)?

To solve for y in this equation, you can use the properties of logarithms to rewrite the equation as y^2=ln(1-y^2)=ln(e^y^2). Then, you can take the natural log of both sides to get y^2=2y^2. Finally, you can solve for y by taking the square root of both sides, giving you y=±√2.

## 2. Is there more than one solution for y in y^2=ln(1-y^2)?

Yes, there are two solutions for y in this equation: y=√2 and y=-√2. This is because when you take the square root of both sides, you get both a positive and negative value for y.

## 3. Can I use a calculator to solve for y in y^2=ln(1-y^2)?

Yes, you can use a calculator to solve for y in this equation. You can use the "ln" button to take the natural log of both sides, and then use the square root function to solve for y.

## 4. Can I use the quadratic formula to solve for y in y^2=ln(1-y^2)?

No, you cannot use the quadratic formula to solve for y in this equation. The quadratic formula is used to solve equations in the form ax^2+bx+c=0, but the equation y^2=ln(1-y^2) is not in this form.

## 5. Can I solve for y if the equation is y^2=ln(1-y)?

Yes, you can still solve for y if the equation is y^2=ln(1-y). You would use the same steps as in the first question, but you would have a different solution: y=±√2. This is because the equation is now y^2=ln(1-y)=ln(e^y), which simplifies to y^2=y.

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