Implicit Differentiation Problem

[steven10137]

Homework Statement

If (1+x^2)(y^2)=1-x^2, show that (\frac{d}{{dx}})^2=\frac{1-y^4}{{1-x^4}}

2. The attempt at a solution

http://img294.imageshack.us/img294/7133/calcqn1qd8.gif

I have got to this point and tried to simplify the problem with no success ...
Have I made an error in my differentiation or is there a way to possibly simplify this equation to the required form??

thanks in advance
Steven

- sorry bout the quality of the image it was the best i could do with my scanner ...

 

[pizzasky]

Your differentiation is correct. Subsequently, start with (\frac{dy}{dx})^2 = \frac{x^2(1+y^2)^2}{y^2(1+x^2)^2} and work from there (do not expand everything).

Eventually, you should be able to show that (\frac{dy}{dx})^2 = \frac{1-y^4}{1-x^4}.

 

[cristo]

I'm not too sure what you mean by this (\frac{d}{{dx}})^2=\frac{1-y^4}{{1-x^4}}. I'd say that you really mean \left(\frac{d}{{dx}}\right)^2y=\frac{d^2y}{dx^2}=\frac{1-y^4}{{1-x^4}}.

In other words, you want to show that the second derivative of y wrt x is (1-y^4)/(1-x^4). To do this, you need to differentiate \frac{dy}{dx}=\frac{x(1+y^2)}{y(x^2+1)}, implicitly.

 

[steven10137]

I'm not too sure what you mean by this (\frac{d}{{dx}})^2=\frac{1-y^4}{{1-x^4}}. I'd say that you really mean \left(\frac{d}{{dx}}\right)^2y=\frac{d^2y}{dx^2}=\frac{1-y^4}{{1-x^4}}.

In other words, you want to show that the second derivative of y wrt x is (1-y^4)/(1-x^4). To do this, you need to differentiate \frac{dy}{dx}=\frac{x(1+y^2)}{y(x^2+1)}, implicitly.

my mistake sorry, i didn't show it clearly enough:
\left(\frac{dy}{{dx}}\right)^2=\frac{d^2y}{dx^2}=\frac{1-y^4}{{1-x^4}}
is what i meant that was a typo.

Just to clarify;
\frac{dy}{dx}=-\frac{x(1+y^2)}{y(x^2+1)}
Note the negative sign ...

Im not sure how I can go about differentiating implicitly:
\frac{dy}{dx}=-\frac{x(1+y^2)}{y(x^2+1)}
If you could give me any suggestions as to how to approach that it would be appreciated

cheers for your fast responses
Steven

 

[Office_Shredder]

To get the second derivative:

Just differentiate

(1+x^2)(y^2)=1-x^2

twice with respect to x. You'll get some y''s and some y's. You know what y' is, so you can sub that in and solve for y'' in terms of y and x.

If you're a bit hesitant to doubly differentiate implicitly, you can take your function y'(x,y) and differentiate that with respect to x. Again, implicitly differentiating the RHS. and the LHS becomes y''. Then plug in your function of x and y for y' into the RHS wherever y' shows up, and you should be done

 

[daveb]

You have your expression for dy/dx. Square that like you did to get but don't expand the numerator and denominator. Instead, solve the initial equation for x^2, and then also for y^2, and see if cleverly substituting these into the result works.

Edit: Ah, second derivative, not square of the first derivative. I haven't worked out the 2nd derivative yet, but oddly the square of the first derivative does work out correctly.

 

[cristo]

my mistake sorry, i didn't show it clearly enough:
\left(\frac{dy}{{dx}}\right)^2=\frac{d^2y}{dx^2}=\frac{1-y^4}{{1-x^4}}
is what i meant that was a typo.

Just to clarify;
\frac{dy}{dx}=-\frac{x(1+y^2)}{y(x^2+1)}
Note the negative sign ...

Im not sure how I can go about differentiating implicitly:
\frac{dy}{dx}=-\frac{x(1+y^2)}{y(x^2+1)}
If you could give me any suggestions as to how to approach that it would be appreciated

cheers for your fast responses
Steven

Ok, well I misinterpreted your original post. You do indeed require the square of dy/dx, and not the second derivative of y wrt x. Note that \left(\frac{dy}{dx}\right)^2 and \frac{d^2y}{dx^2} are not the same thing!

You have found dy/dx, and so now need to square it. Then follow the advice given above, and use the original equation to eliminate x^2 and y^2 in your expression for (y')^2. This should yield the required expression, after a little cancellation.

 

[Office_Shredder]

If you really meant

\left(\frac{dy}{{dx}}\right)^2=\frac{d^2y}{dx^2}=\ frac{1-y^4}{{1-x^4}}

then you want the square of the first derivative.