Exercise · ZW-GRADIENT-004 · ★★☆ Intermediate

Compute \(\nabla f(1,1)\) for \(f(x,y)=\ln(x^2+y^2)\).

Choose one option.

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  • B. \(\langle 1,1\rangle\)

Explanation

\(f_x=\frac{2x}{x^2+y^2}=1\) and \(f_y=\frac{2y}{x^2+y^2}=1\) at \((1,1)\). Forgetting the denominator would give \(\langle2,2\rangle\).

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Review the concept: Gradient

The gradient \(\nabla f\) of a differentiable function \(f\) is the vector that collects its partial derivatives at a point; it points where \(f\) increases fastest.

\[\nabla f=\langle f_x,\,f_y\rangle=f_x\,\mathbf{i}+f_y\,\mathbf{j}\]

More practice problems: gradient

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