Partial Derivatives and Differentiability

Freeze the other coordinates

The partial derivative $f_x(a,b)$ is the one-variable derivative of $f(x,b)$ at $a$ . The partial $f_y$ holds $x$ fixed instead. Higher partials record repeated coordinate differentiation. If mixed second partials are continuous near a point, Clairaut’s theorem gives $f_{xy}=f_{yx}$ there.

Differentiability is stronger than the mere existence of partial derivatives. It requires

f(a+h,b+k)=f(a,b)+f_x(a,b)h+f_y(a,b)k+o(\sqrt{h^2+k^2}).

Continuous first partials in a neighborhood are a useful sufficient condition. They are not a necessary condition for every differentiable function.

Visual guide

VISUAL GUIDEA partial derivative follows a coordinate slice

Slice y = 2
Tangent z = 3 + 4(x − 1)

For f(x, y) = x² + xy at (1, 2), fixing y = 2 gives the blue slice x² + 2x. Its tangent slope at x = 1 is 4. Fixing x instead would measure the other partial derivative, not this slope.

Worked example: a tangent-plane estimate

For $f(x,y)=x^2+xy$ at $(1,2)$ , $f=3$ , $f_x=2x+y=4$ and $f_y=x=1$ . The tangent plane is $z=3+4(x-1)+(y-2)$ . At $(1.02,1.97)$ it predicts $3.05$ .

The exact value is $1.02^2+1.02(1.97)=3.0498$ . The error is $-0.0002$ , arising from the quadratic terms $h^2+hk$ . A local plane can be accurate without agreeing exactly with the surface.

Worked example: partials are not enough

Define $f(x,y)=xy/(x^2+y^2)$ away from zero and $f(0,0)=0$ . Along each coordinate axis the function is zero, so both partial derivatives at the origin exist and equal zero. Along $y=x$ , however, the values stay at $1/2$ . The function is not continuous and therefore cannot be differentiable there.

A formula for a plane constructed from partials does not establish that the plane is tangent in the differentiability sense. Check sufficient regularity or the remainder definition when the point is exceptional.

Practice

Find both first partials of $x^2y+\sin y$ .
Linearize $f=x^2+y^2$ at $(1,1)$ .
Compute $f_{xy}$ and $f_{yx}$ for $f=x^2y^3$ .

Show worked solutions

$f_x=2xy$ and $f_y=x^2+\cos y$ .
$L=2+2(x-1)+2(y-1)$ .
Both equal $6xy^2$ . Polynomial partials are continuous, so equality also follows from the theorem.

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.