Find the derivative from the equation?

Question

Please help me find dy/dx of the
arctan (y/x) = 1/2 * ln(x^2+y^2)
I don't know what kind of this function is to find the derivative. Is it implicit function?

Answer 1

The nicest way to do this is with polar coordinates.

Note that in polar coordinates the equation becomes
θ = ln r
i.e. r = e^θ
So x = r cos θ = e^θ cos θ and y = r sin θ = e^θ sin θ.
So dx/dθ = e^θ cos θ - e^θ sin θ = x - y
and dy/dθ = e^θ sin θ + e^θ cos θ = y + x
and by the chain rule
dy/dx = dy/dθ . dθ/dx
= dy/dθ / dx/dθ
= (x+y) / (x-y).

Alternatively, we can use implicit differentiation as follows:

arctan (y/x) = (1/2) ln(x^2+y^2)
<=> y/x = tan ((1/2) ln(x^2+y^2))
<=> y = x tan ((1/2) ln(x^2+y^2))
dy/dx = tan ((1/2) ln(x^2+y^2)) + x (sec^2 ((1/2) ln(x^2+y^2))) . (1/2) (1/(x^2 + y^2)) (2x + 2y dy/dx)
Now, note that sec^2 ((1/2) ln(x^2+y^2)) = 1 + tan^2 ((1/2) ln(x^2+y^2)) = 1 + (y/x)^2 = (x^2 + y^2) / x^2. So we get

dy/dx = y/x+ x ((x^2 + y^2) / x^2) . (1/(x^2 + y^2)) (x + y dy/dx)
= y/x + (1 / x) . (x + y dy/dx)
= y/x + 1 + y/x dy/dx
<=> dy/dx (1 - y/x) = y/x + 1
<=> dy/dx = (y/x + 1) / (1 - y/x)
= (x + y) / (x - y).

Answer 2

You would use implicit differentiation. It may be easier to take tan of both sides to get

y/x = tan[1/2*ln(x^2 + y^2)]

since you are more likely to know the derivative of tan(x) than arctan(x).

Now differentiate both sides with respect to x:

-y/x^2 + y'/x = sec^2[1/2*ln(x^2 + y^2)] * d/dx [1/2*ln(x^2 + y^2)]

etc (continue with differentiation of the right side).

Answer 3

D[arctan (x)]=1/(1+x²)dx. So:

arctan (y/x) = 1/2 * ln(x²+y²)
[1/(1+y²/x²)] (xdy-ydx)/x² =(xdx+ydy)/(x²+y²)
1/[(x²+y²)/x²] (xdy-ydx)/x² =(xdx+ydy)/(x²+y²)
x²/(x²+y²) (xdy-ydx)/x² =(xdx+ydy)/(x²+y²)

Simplify x² & (x²+y²):
xdy-ydx =xdx+ydy
xdy-ydy=xdx+ydx
(x-y)dy=(x+y)dx
dy/dx=(x+y)/(x-y)

I hope this help. Bye.