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Why is the existence of two partial derivatives alone insufficient for a function to be differentiable?

Differentiability requires that the linear approximation (the tangent plane) becomes arbitrarily accurate as the input displacement approaches zero. Specifically, the remaining error between the actual increment and the total differential, divided by the input displacement length, must tend to zero. Merely having partial derivatives does not guarantee this limit condition is met.

Conditions

  • The function is evaluated at a reference point (x0,y0)(x_0, y_0).
  • The input is displaced by (Δx,Δy)(\Delta x, \Delta y) with length ρ=Δx2+Δy2\rho = \sqrt{\Delta x^2 + \Delta y^2}.

Reasoning, step by step

  1. Recall the definition of differentiability: the error Δz−dz\Delta z - dz must be o(ρ)o(\rho) as ρ→0\rho \to 0.
  2. Understand that this means lim⁡ρ→0Δz−dzρ=0\lim_{\rho \to 0} \frac{\Delta z - dz}{\rho} = 0.
  3. Recognize that the existence of partial derivatives only guarantees the slopes in the x and y directions.
  4. Conclude that without the limit condition on the error, the surface might not have a well-defined tangent plane that approximates it in all directions.

Example

The script states: 'Differentiability means the remaining error divided by the input displacement length tends to zero. Existence of the two partial derivatives alone is insufficient.' The card formula is Δz−dz=o(Δx2+Δy2)\Delta z-dz=o(\sqrt{\Delta x^2+\Delta y^2}).

Common misconceptions

  • Believing that if partial derivatives exist, the function is automatically differentiable.
  • Thinking that the tangent plane is defined solely by the partial derivatives without requiring the error limit condition.
  • Assuming that continuity of the function is sufficient for differentiability.

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