Directional derivative of ln(x^2 + y^2) at (1, 2)

Log outside, chain rule inside

For $\ln(u)$ the partial is $u_x / u$, so the inner derivative sits on top. With $u = x^2 + y^2$ that gives $2x$ over $u$ in the $x$ slot and $2y$ over $u$ in the $y$ slot.

At $(1,2)$ the denominator is $1 + 4 = 5$. Dotting with $\mathbf{u} = \langle 0.8, -0.6 \rangle$ takes one line.

The mistake: dropping the inner derivative

The usual slip is writing $\frac{\partial f}{\partial x} = \frac{1}{x^2+y^2}$, as though $\ln$ of a lump differentiates like $\ln x$. That halves the $x$ component here and quarters the $y$ component, giving $\frac{1}{5}(0.8) + \frac{1}{5}(-0.6) = 0.04$, a positive number where the truth is negative.

A sign check catches it. The gradient always points away from the origin here, because $\ln(x^2+y^2)$ grows with distance, so the sign of the answer is the sign of the direction's component along the outward radial vector $\langle 1, 2 \rangle / \sqrt{5}$. For $\langle 0.8, -0.6 \rangle$ that component is $(0.8 - 1.2)/\sqrt{5} \approx -0.179$, pointing inward, so a negative answer is what you should expect. A negative $y$ part on its own proves nothing: $\langle 0.96, -0.28 \rangle$ also tilts downward and gives $+0.16$.

Reading the negative sign

The level curves of $f$ are circles centred at the origin, and the gradient is perpendicular to them. The angle between $\nabla f(1,2)$ and $\mathbf{u}$ is about 100.3 degrees, just past the right angle that would give zero.

• Angle under 90 degrees: $f$ increases along that direction
• Angle exactly 90 degrees: you are moving along the level circle and the rate is 0
• Angle over 90 degrees, as here: $f$ decreases, and the value is negative

The size is small because $\lVert \nabla f(1,2) \rVert = \frac{2}{\sqrt{5}} \approx 0.894$ and the cosine is only about $-0.179$.