Directional derivative of e^(2x - y) at (1, 2)

One exponential, two chain factors

The exponential reproduces itself and each partial multiplies by the partial of the exponent. Here the exponent $2x - y$ contributes $2$ in the $x$ slot and $-1$ in the $y$ slot.

At $(1,2)$ the exponent is $2(1) - 2 = 0$, so the common factor $e^{0}$ is 1 and the gradient is simply $\langle 2, -1 \rangle$. Choosing a point on the line $y = 2x$ is what makes the arithmetic clean.

The mistake: losing the 2 or the minus

Writing $\frac{\partial f}{\partial x} = e^{2x-y}$ without the 2 gives a gradient of $\langle 1, -1 \rangle$ and an answer of $-1.4$. Writing $\frac{\partial f}{\partial y} = e^{2x-y}$ without the minus gives $\langle 2, 1 \rangle$ and an answer of $-1.0$. Neither is close.

There is also a reading mistake. A negative directional derivative does not mean the function is negative; $e^{2x-y}$ is positive everywhere. It means the value is falling as you move that way, from 1 toward smaller positive numbers.

You can see why from the exponent alone. Moving along $\langle -0.8, 0.6 \rangle$ decreases $2x$ and increases $y$, so $2x - y$ drops and the exponential follows it down.

Close to the steepest descent

The fastest decrease at a point runs along $-\nabla f$, at rate $-\lVert \nabla f \rVert$. Here that is $-\sqrt{5} \approx -2.2361$, in the direction $\langle -2, 1 \rangle / \sqrt{5} \approx \langle -0.894, 0.447 \rangle$.

• Our direction $\langle -0.8, 0.6 \rangle$ is about 10.3 degrees off that steepest descent direction
• The rate $-2.2$ is therefore about 98.4 percent of the fastest possible drop
• Any direction perpendicular to $\langle 2, -1 \rangle$, such as $\langle 1, 2 \rangle / \sqrt{5}$, gives exactly 0

The level curves are the lines $2x - y = c$, which is why a whole family of points shares the same behaviour along a fixed direction.