Approximation

If f(x,y)f(x,y) is differentiable at (x0,y0)(x_0,y_0) then:

f(x,y)≈f(x0,y0)+fx(x0,y0) Δx+fy(x0,y0) Δyf(x,y) \approx f(x_0,y_0) + f_x(x_0,y_0)\,\Delta x + f_y(x_0,y_0)\,\Delta y

Total Differential

For z=f(x,y)z = f(x,y), the total differential is:

dz=fx(x,y) dx+fy(x,y) dydz = f_x(x,y)\,\text{d}x + f_y(x,y)\,\text{d}y

dzdz gives the exact change in the tangent plane for increments dxdx and dydy. It approximates Δz=f(x+dx,y+dy)−f(x,y)\Delta z = f(x+dx, y+dy) - f(x,y) when dxdx and dydy are small.

Example

Approximate f(2.01, 3.98)f(2.01,\, 3.98) where f(x,y)=x2+xyf(x,y) = x^2 + xy.

Solution Reveal Hide

At (x0,y0)=(2,4)(x_0, y_0) = (2, 4):

  • f(2,4)=4+8=12f(2,4) = 4 + 8 = 12
  • fx=2x+y⇒fx(2,4)=8f_x = 2x + y \Rightarrow f_x(2,4) = 8
  • fy=x⇒fy(2,4)=2f_y = x \Rightarrow f_y(2,4) = 2
Δf≈8(0.01)+2(−0.02)=0.08−0.04=0.04\Delta f \approx 8(0.01) + 2(-0.02) = 0.08 - 0.04 = 0.04f(2.01, 3.98)≈12.04f(2.01,\, 3.98) \approx 12.04
Written by October 19, 2026 1 min read
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