Find the tangenplane at (1,2,3) $f (x, y ) = \frac{x^2y}{y − 1} + 1$, then approximate the value of the function at f(1.2, 2.3)
I partial differentiated and got $f_x$ = 4 and $f_y=-1$ then I used the formula for tangent plane $L(x,y) = f(x,y) + f_x(a,b)(x-x_0) + f_y(a,b) (y-y_0)$ and I got $f(x,y) = 3 + 4(1.2-1) - (2.3-2) = 3 + 0.8 - 0.3 = 3.5$ but the correct answer is $2.5$.
What am I doing wrong?
Just to give a reply, your solution is fine. You can check by yourself that if you apply $f(1.2,2.3)$ then you get approx, $3.5476$ which is compatible with your answer instead of $2.5$.