implicit function x cos(xy) = 0 defines y as fcn of x in....

[xoninhas]

Well again I'm having problems with this...

"Show that the equation x*cos(x*y) = 0 defines, implicitly, y as a function of x in some neighborhood of (1,pi/2) and calculate the derivative df(1)/dx."

Well supposedly I should use the implicit function theorem since it is not possible to isolate the y.
For that F(x,y) = x*cos(x*y) = 0 and det(DF(1,pi/2)/dx) != 0.

Well so how is the determinant of a scalar??! And in fact in that it IS 0.... What am I doing wrong??

Thank you again!

 

[tkhunny]

I'm not seeing the problem with this one. This appears to be a very nicely behaved relation at the indicated location.

Why do you need a determinant? Why isn't old fashioned implicit differentiation sufficient?

 

[xoninhas]

Can you elaborate more on the implicit differentiation? What is that? Cause from what I've seen it's always determinant envolved...

EDIT: Progress report

Well I did the derivative of df(x,y)/dy = x*sen(x*pi/2) which in the point (1, pi/2) is different from zero (-1) So I think that means that there is a y = f(x) <=> F(x,y) = 0.

Now for the derivative I'm a bit more confused... any pointers?

 

[tkhunny]

Assuming \(\displaystyle y = f(x)\), we have:

\(\displaystyle x*\cos(x*y) = x*\cos(x*f(x)) = 0\)

Then \(\displaystyle (d/dx)[x*\cos(x*f(x))] = x[-\sin(x*f(x))*[x*f'(x)+f(x)]] + \cos(x*f(x)) = 0\)

That is not difficult to solve for \(\displaystyle f'(x)\) and it is seen that there is no difficulty around \(\displaystyle (1,\frac{\pi}{2})\).

 

[xoninhas]

Thank you so much! Can I just check with you guys to know if the result is -pi/2 ?

Thanks again!

 

[galactus]

That's what I get.