Partial derivatives and total differentiability

Try

Use f(x,y)=xy+y² at the point (2,1). Its value is 3. If x changes by a small amount while y stays 1, the change is exactly the change in x. If y changes while x stays 2, both terms contribute. A partial derivative fixes the other independent coordinates; it is not the rate along an arbitrary path where several coordinates change together. State what is being held constant before differentiating.

Predict

Predict the change when x increases by 0.02 and y decreases by 0.01. Here f_x=y=1 and f_y=x+2y=4 at the base point. The linear prediction is 1(0.02)+4(−0.01)=−0.02, so the predicted new value is 2.98. Direct evaluation gives 2.9799. The discrepancy is small, but its size needs an explanation rather than the vague claim that linear approximation is always accurate.

Experiment

Write the increments as h and k. Expanding exactly gives f(2+h,1+k)=3+h+4k+hk+k². The remainder is hk+k². Use |hk|≤(h²+k²)/2 and k²≤h²+k² to bound its magnitude by 3(h²+k²)/2. Dividing by the displacement length √(h²+k²) leaves a bound tending to zero. Try halving both increments: the linear part halves, while the remainder quarters. This is the mechanism behind local linear behavior.

Understand

Total differentiability at p means there is a linear map L such that f(p+h)=f(p)+L(h)+o(|h|). The notation o(|h|) means the error divided by |h| tends to zero as the displacement tends to zero from every direction. For a scalar function of two variables, L(h,k)=f_x(p)h+f_y(p)k when differentiability holds.

The tangent plane is z=f(a,b)+f_x(a,b)(x−a)+f_y(a,b)(y−b). It describes the first-order surface approximation, not a global replacement for the graph. Continuous first partials on a neighborhood are a standard sufficient condition for differentiability. Merely possessing two partial derivatives at one point is insufficient because those derivatives inspect only two coordinate directions.

Apply

Challenge the invalid proof “the axis slopes are zero, so the tangent plane z=0 approximates everything.” Define q(0,0)=0 and q(x,y)=xy/√(x²+y²) elsewhere. Both partial derivatives at the origin are zero, and the function is continuous there. Yet along y=x>0, q=x/√2 and the displacement norm is √2x. Their ratio is 1/2, not a quantity tending to zero. The function is not differentiable at the origin despite being continuous and having partials.

For smooth functions, second partials measure changes in the first partials. In this worked polynomial, f_xy=f_yx=1. Equality of mixed partials has hypotheses: continuity of the relevant second partials near the point is a standard sufficient condition, not an automatic property of every twice partially differentiable function.

Continue

For optional depth, express the total differential as df=f_x dx+f_y dy and interpret dx,dy as input perturbations. This notation is useful for sensitivity estimates, but it does not erase the remainder or certify a finite-error bound without further information. If inputs have different physical units, each coefficient must convert its own input unit to the output unit.

Use the partial derivative slice investigation to compare coordinate slices. Then explain what those slices leave untested. Keep the exact polynomial remainder as evidence of differentiability, and continue to the chain rule, which combines these local linear maps along a moving path. A numerical estimate may support intuition, but the all-direction remainder argument is what establishes the differentiability claim.