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Partial Derivatives & Directional Change

Useful preparation: Derivatives, Chain Rule & Implicit Change

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Meaning & notation

Partial derivatives measure changes along coordinate slices. For a differentiable function, the gradient combines these rates into a local linear model for every direction. Directional change is its projection onto a unit direction.

Partial derivative
The instantaneous rate when one input varies and the others remain fixed; subscripts identify the varying coordinate.
Gradient
The vector of first partial derivatives. For a differentiable scalar function it encodes the linear part of all small input changes.
Directional derivative
The rate of change per unit distance in a specified direction. A unit vector sets direction without introducing an extra speed factor.

Conditions before calculation

  • The gradient dot-product formula requires differentiability at the point. Continuous first partial derivatives in a neighborhood are a sufficient condition.
  • Normalize a nonzero direction vector before asking for change per unit distance; its unnormalized version instead includes its magnitude as a speed factor.
  • The gradient identifies a unique steepest-ascent direction only when it is nonzero. Orthogonality to a regular level curve also assumes a nonzero gradient.

Why the method works

1. Differentiate coordinate slices

Vary one input while treating the others as constants. Assemble these coordinate derivatives into a vector at the specified point.

∇f(a,b)=⟨fx(a,b),fy(a,b)⟩\nabla f(a,b)=\langle f_x(a,b),f_y(a,b)\rangle

2. Use the local linear model

Differentiability provides one linear approximation for every small displacement, with error negligible relative to displacement length. Merely existing coordinate partials are insufficient.

f(p+h)=f(p)+∇f(p)⋅h+o(∥h∥),Duf(p)=∇f(p)⋅uf(\mathbf p+\mathbf h)=f(\mathbf p)+\nabla f(\mathbf p)\cdot\mathbf h+o(\|\mathbf h\|),\qquad D_{\mathbf u}f(\mathbf p)=\nabla f(\mathbf p)\cdot\mathbf u

3. Read geometry from the dot product

The projection is largest along a nonzero gradient and vanishes perpendicular to it. Zero first-order change does not imply constancy along a straight path.

∥u∥=1,∇f(p)≠0:umax⁡=∇f(p)∥∇f(p)∥,max⁡Duf=∥∇f(p)∥\|\mathbf u\|=1,\quad \nabla f(\mathbf p)\ne\mathbf0:\qquad \mathbf u_{\max}=\frac{\nabla f(\mathbf p)}{\|\nabla f(\mathbf p)\|},\quad \max D_{\mathbf u}f=\|\nabla f(\mathbf p)\|

Worked examples

EXAMPLE 1

Find the directional derivative at the given point toward the vector.

  1. Compute the coordinate rates before substituting the point. Polynomial smoothness justifies the gradient formula.

    f(x,y)=x2y+3y2,p=(1,2);fx=2xy,fy=x2+6y,∇f(1,2)=⟨4,13⟩f(x,y)=x^2y+3y^2,\quad \mathbf p=(1,2);\qquad f_x=2xy,\quad f_y=x^2+6y,\quad \nabla f(1,2)=\langle4,13\rangle

  2. The direction vector is not a unit vector. Divide by its length to separate direction from speed.

    v=⟨3,4⟩,∥v∥=5,u=⟨35,45⟩\mathbf v=\langle3,4\rangle,\qquad \|\mathbf v\|=5,\qquad \mathbf u=\left\langle\frac35,\frac45\right\rangle

  3. Take the projection. Compare it with the gradient magnitude, the largest possible rate at this point.

    Duf(1,2)=435+1345=645,max⁡Duf(1,2)=185D_{\mathbf u}f(1,2)=4\frac35+13\frac45=\frac{64}{5},\qquad \max D_{\mathbf u}f(1,2)=\sqrt{185}

Conclusion: The requested rate is 12.8 per unit distance, below the steepest possible rate.

EXAMPLE 2

Can a direction with zero directional derivative still change the function value?

  1. At the specified point, compute the gradient and choose a perpendicular unit direction.

    T(x,y)=10−x2−4y2,p=(1,1),∇T(p)=⟨−2,−8⟩,u=⟨4,−1⟩17T(x,y)=10-x^2-4y^2,\quad \mathbf p=(1,1),\quad \nabla T(\mathbf p)=\langle-2,-8\rangle,\quad \mathbf u=\frac{\langle4,-1\rangle}{\sqrt{17}}

  2. The dot product vanishes, so the initial first-order rate along the straight path is zero.

    DuT(p)=−8+817=0,r(s)=p+suD_{\mathbf u}T(\mathbf p)=\frac{-8+8}{\sqrt{17}}=0,\qquad \mathbf r(s)=\mathbf p+s\mathbf u

  3. Substitute the entire path into the function. Linear terms cancel, but the quadratic change remains for every nonzero step.

    T(r(s))=10−(1+4s17)2−4(1−s17)2=5−2017s2T(\mathbf r(s))=10-\left(1+\frac{4s}{\sqrt{17}}\right)^2-4\left(1-\frac{s}{\sqrt{17}}\right)^2=5-\frac{20}{17}s^2

Conclusion: The value decreases for every nonzero step, despite a zero initial directional derivative.

Pitfalls & counterexamples

Tempting claim: Existing partial derivatives guarantee differentiability.

Correction: Coordinate slices alone do not control all approaches to a point.

Test the claim: Away from the origin, divide the coordinate product by the sum of coordinate squares; assign zero at the origin. Both origin partials vanish, but diagonal values equal one half, contradicting continuity.

Check your understanding

What happens to the gradient dot product if a direction vector is doubled without normalization?

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Compare your reasoning

It doubles.

The dot product is linear in the vector; the result now measures change along a path moving twice as fast.

Connect this idea

Gradients supply tangent planes, optimization conditions, and conservative vector fields. The local linear model generalizes to Jacobian matrices for vector-valued functions.

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Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

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