Multivariable calculus

Directional derivative of e^(xy) at (1, 1)

Q: Why is the gradient (e, e) and not (1, 1)?

Each partial is the exponent's partial times e^{xy}. At (1,1) the exponent partials are y = 1 and x = 1, and e^{xy} = e^{1} = e, so both components equal e \approx 2.718.

Q: Is (0.28, 0.96) really a unit vector?

Yes. 0.28^2 + 0.96^2 = 0.0784 + 0.9216 = 1. It is the 7-24-25 right triangle written as a unit vector, and the build re-checks its length on every deploy.

At (1, 1) the gradient of f(x,y) = e^(xy) is (e, e), because each partial picks up a chain rule factor from the exponent. Dotting with the unit direction (0.28, 0.96) gives 1.24e, that is 31e/25 or about 3.3707, the rate of change of f per unit distance in that direction.

f(x,y) = e^{xy}

D_{\mathbf{u}} f(1,1) = \frac{31e}{25} \approx 3.3707

At x = 1, y = 1, in the unit direction (0.28, 0.96).

Both partials come from the chain rule

The outer function is $\exp$ , whose derivative is itself, so each partial is $e^{xy}$ times the partial of the exponent. The exponent $xy$ has $\frac{\partial}{\partial x}(xy) = y$ and $\frac{\partial}{\partial y}(xy) = x$ .

\nabla f = \left\langle y\,e^{xy},\; x\,e^{xy} \right\rangle, \qquad \nabla f(1,1) = \langle e, e \rangle

Now dot with the given unit direction. Keeping $e$ symbolic to the end is cleaner than rounding early, and it makes the fraction obvious.

D_{\mathbf{u}} f(1,1) = e(0.28) + e(0.96) = 1.24e = \frac{31e}{25}

The mistake: forgetting the factor in front

Students who remember only that the exponential is its own derivative write $\frac{\partial f}{\partial x} = e^{xy}$ . At $(1,1)$ that happens to give $e$ as well, so the error hides at this point and shows up everywhere else.

At $(2,1)$ the correct $\frac{\partial f}{\partial x}$ is $1 \cdot e^{2} \approx 7.389$ , and the careless version gives the same thing again
At $(1,2)$ the correct value is $2e^{2} \approx 14.78$ , while the careless version gives $e^{2} \approx 7.389$ , off by a factor of 2
Check a case where the missing factor is not 1, and the mistake becomes visible immediately

The second trap is dotting with $\langle 7, 24 \rangle$ rather than $\langle 0.28, 0.96 \rangle$ . Those are the same direction, since $7^2 + 24^2 = 25^2$ , but the answer would be 25 times too big.

Comparing with the steepest direction

Because the gradient is $\langle e, e \rangle$ , the steepest ascent direction at $(1,1)$ is the 45 degree line $\langle 1, 1 \rangle / \sqrt{2}$ , and the fastest possible rate is $e\sqrt{2} \approx 3.844$ .

D_{\mathbf{u}} f(1,1) = e\sqrt{2}\,\cos\theta, \qquad \theta \approx 28.7^{\circ}

Our direction is about 28.7 degrees off the steepest one, which costs only about 12 percent of the rate. Near the maximum the cosine is flat, so being a little off the best direction barely matters.

The build re-checks that (0.28, 0.96) has length 1, rebuilds the gradient of e^(xy) at (1, 1) by central differences on the function itself, and fails unless the dot product is still 3.3707 to within a relative 1e-4.

Frequently asked questions

Why is the gradient (e, e) and not (1, 1)?

Each partial is the exponent's partial times $e^{xy}$ . At $(1,1)$ the exponent partials are $y = 1$ and $x = 1$ , and $e^{xy} = e^{1} = e$ , so both components equal $e \approx 2.718$ .

Is (0.28, 0.96) really a unit vector?

Yes. $0.28^2 + 0.96^2 = 0.0784 + 0.9216 = 1$ . It is the 7-24-25 right triangle written as a unit vector, and the build re-checks its length on every deploy.