Practice: Partial Derivatives, the Gradient and Critical Points

(a) The gradient. For $f_x$, hold $y$ constant: $x^{2}y$ differentiates to $2xy$, and $3y^2$ contains no $x$, so it contributes 0. For $f_y$, hold $x$ constant: $x^{2}y$ differentiates to $x^2$, and $3y^2$ to $6y$.

Check: symmetric difference quotients at $h={10}^{-6}$ give $4.00000000$ and $10.00000000$ ✓. Note $f_x=2xy$ still contains $y$, holding a variable fixed means treating it as a constant while differentiating, not removing it from the result. (b) A directional derivative, and the steepest direction. The vector $(3,4)$ has length $\sqrt{9+16}=5$, so it is not a unit vector and must be normalised:

Check: $\frac{f(p+tu)-f(p-tu)}{2t}$ at $t={10}^{-6}$ returns $10.40000000$ ✓. Without normalising the dot product gives $4(3)+10(4)=52$, exactly 5 times too large, since the vector is 5 times too long. A directional derivative is a rate per unit distance, so the length must be 1. Steepest increase. By $D_{u}f=|\mathrm{\nabla }f|\cos \theta$, the maximum is at $\theta =0$: along $\mathrm{\nabla }f=(4,10)$ itself, at rate

As a unit vector that direction is $(4,10)/\sqrt{116}\approx (0.3714,0.9285)$, and dotting $\mathrm{\nabla }f$ with it returns $10.770$, equal to the magnitude ✓. Scanning all directions at a hundredth of a degree finds no direction beating it. Note , as it must be: no directional derivative can exceed the gradient's length. That inequality is the quickest check on any answer here. (c) Critical points of $g(x,y)=x^2+3y^2-4x+6y$. Both partials must vanish simultaneously:

Each equation alone describes a line; their intersection is the single critical point $(2,-1)$, where

Classification. $g_{xx}=2$, $g_{yy}=6$, $g_{xy}=0$, so

and . Therefore $(2,-1)$ is a local minimum, value $-7$. Check: 20,000 random points within $0.5$ of $(2,-1)$ produced no value below $-7$ ✓. (d) The origin as a saddle for $s(x,y)=x^2-y^2$. $s_x=2x$ and $s_y=-2y$ vanish together only at $(0,0)$, where $s=0$. Explicit values. Along the $x$-axis:

both greater than $s(0,0)=0$. Along the $y$-axis:

both less than 0. Every neighbourhood of the origin therefore contains points where $s$ is larger and points where it is smaller, so the origin is neither a maximum nor a minimum. By the test. $s_{xx}=2$, $s_{yy}=-2$, $s_{xy}=0$, giving , which is precisely the saddle case. Note that alone would suggest a minimum. A single second derivative describes one direction; the determinant is what ranges over all of them.