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How is the total differential dz of a differentiable two-variable function computed from its partial derivatives at a reference point?

The total differential dzdz is computed by evaluating both partial derivatives at the reference point (x0,y0)(x_0, y_0) and taking their linear combination with the input displacements Δx\Delta x and Δy\Delta y. The formula is dz=fx(x0,y0)Δx+fy(x0,y0)Δydz = f_x(x_0, y_0)\Delta x + f_y(x_0, y_0)\Delta y. This represents the height change on the tangent plane for the same input displacement.

Conditions

  • The function is differentiable at the reference point (x0,y0)(x_0, y_0).
  • The input is moved by a displacement (Δx,Δy)(\Delta x, \Delta y).

Reasoning, step by step

  1. Identify the reference point (x0,y0)(x_0, y_0) and the input displacement (Δx,Δy)(\Delta x, \Delta y).
  2. Evaluate the partial derivative with respect to xx at the reference point: fx(x0,y0)f_x(x_0, y_0).
  3. Evaluate the partial derivative with respect to yy at the reference point: fy(x0,y0)f_y(x_0, y_0).
  4. Multiply each partial derivative by its corresponding input displacement.
  5. Sum the products to obtain the total differential: dz=fx(x0,y0)Δx+fy(x0,y0)Δydz = f_x(x_0, y_0)\Delta x + f_y(x_0, y_0)\Delta y.

Example

The script states: 'Evaluate both partial derivatives at the reference point: dz=fx(x0,y0)Δx+fy(x0,y0)Δydz=f_x(x_0,y_0)\Delta x+f_y(x_0,y_0)\Delta y. The total differential dz is the tangent plane’s height change for the same input displacement.'

Common misconceptions

  • Believing that the existence of the two partial derivatives alone is sufficient to compute the total differential; differentiability is required.
  • Confusing the total differential dzdz (tangent plane height change) with the actual increment Δz\Delta z (actual surface height change).
  • Thinking that the partial derivatives should be evaluated at the new point (x0+Δx,y0+Δy)(x_0+\Delta x, y_0+\Delta y) rather than the reference point.

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