Question

Find the first order partial derivatives of \(f(x, y) = \frac{3xy}{(y - 1)^2}\) \(f_x(x, y) = \frac{3y}{(y - 1)^2}\) and \(f_y(x, y) = \frac{-3x}{(y - 1)^3}\) \(f_x(x, y) = \frac{3y}{(y - 1)^2}\) and \(f_y(x, y) = \frac{3x}{-2(y - 1)}\) \(f_x(x, y) = \frac{3y^2}{(y - 1)^2}\) and \(f_y(x, y) = \frac{-3x(y + 1)}{(y - 1)^3}\) \(f_x(x, y) = \frac{3y}{(y - 1)^2}\) and \(f_y(x, y) = \frac{3x}{(y - 1)^3}\) \(f_x(x, y) = \frac{3}{(y - 1)^2}\) and \(f_y(x, y) = \frac{3x}{2(y - 1)}\) \(f_x(x, y) = \frac{3y}{(y - 1)^2}\) and \(f_y(x, y) = \frac{-3x(y + 1)}{(y - 1)^3}\)

          Find the first order partial derivatives of
\(f(x, y) = \frac{3xy}{(y - 1)^2}\)
\(f_x(x, y) = \frac{3y}{(y - 1)^2}\) and \(f_y(x, y) = \frac{-3x}{(y - 1)^3}\)
\(f_x(x, y) = \frac{3y}{(y - 1)^2}\) and \(f_y(x, y) = \frac{3x}{-2(y - 1)}\)
\(f_x(x, y) = \frac{3y^2}{(y - 1)^2}\) and \(f_y(x, y) = \frac{-3x(y + 1)}{(y - 1)^3}\)
\(f_x(x, y) = \frac{3y}{(y - 1)^2}\) and \(f_y(x, y) = \frac{3x}{(y - 1)^3}\)
\(f_x(x, y) = \frac{3}{(y - 1)^2}\) and \(f_y(x, y) = \frac{3x}{2(y - 1)}\)
\(f_x(x, y) = \frac{3y}{(y - 1)^2}\) and \(f_y(x, y) = \frac{-3x(y + 1)}{(y - 1)^3}\)

Find the first order partial derivatives of
f(x, y) = (3xy)/((y - 1)^2)
fx(x, y) = (3y)/((y - 1)^2) and fy(x, y) = (-3x)/((y - 1)^3)
fx(x, y) = (3y)/((y - 1)^2) and fy(x, y) = (3x)/(-2(y - 1))
fx(x, y) = (3y^2)/((y - 1)^2) and fy(x, y) = (-3x(y + 1))/((y - 1)^3)
fx(x, y) = (3y)/((y - 1)^2) and fy(x, y) = (3x)/((y - 1)^3)
fx(x, y) = (3)/((y - 1)^2) and fy(x, y) = (3x)/(2(y - 1))
fx(x, y) = (3y)/((y - 1)^2) and fy(x, y) = (-3x(y + 1))/((y - 1)^3)

Added by Chris H.

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