Partial Derivatives

A partial derivative measures how a function changes when one input varies and the others stay fixed. It gives a local rate along a coordinate direction, not automatically a complete description of combined input changes.

That distinction matters when building a tangent plane, estimating a finite change or interchanging mixed derivatives. Useful formulas need their hypotheses, and an approximation must not be reported as an exact difference.

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Keep the fixed input visible

For a function of two variables,

f_x(a,b)=\lim_{h\to0}\frac{f(a+h,b)-f(a,b)}h,

f_y(a,b)=\lim_{k\to0}\frac{f(a,b+k)-f(a,b)}k.

The second input remains fixed in the first limit; the first remains fixed in the second. At an interior point these are ordinary two-sided limits along coordinate lines.

If the function is instead restricted to a path where both inputs change, the total derivative generally has additional contributions. It is a different operation, even when similar symbols appear.

Differentiate a two-variable polynomial

Let

f(x,y)=x^2y+3xy^2.

Holding $y$ constant gives $f_x=2xy+3y^2$ . Holding $x$ constant gives $f_y=x^2+6xy$ . At $(1,2)$ ,

f_x(1,2)=16,\qquad f_y(1,2)=13.

A term independent of the differentiated input has derivative zero, even when it depends on another input. A fixed-input factor acts as a constant coefficient during that calculation.

Interpret partials as slopes of traces

Fixing one input slices a surface into a single-variable curve. The partial derivative is the slope of that trace. Two coordinate traces do not by themselves establish that one plane approximates the surface well in every direction.

Two trace changes at (1,2)

The mathematical inputs are dimensionless in this example.

Diagram explanation and exact data

The partial slopes for g are two and twelve, distinct from the earlier polynomial f in the same lesson.

Two trace changes at (1,2): exact reference data
h	x-only change	y-only change
−0.1	−0.19	−1.17
0	0	0
0.1	0.21	1.23

The diagram uses the separate function $g(x,y)=x^2+3y^2$ at $(1,2)$ , developed below. Its partial rates are two and twelve. They must not be mixed with sixteen and thirteen from the earlier polynomial example.

Differentiability is a complete linear approximation

A function is differentiable at $(a,b)$ when

f(a+h,b+k)=f(a,b)+f_x(a,b)h+f_y(a,b)k+R(h,k),

where

\frac{R(h,k)}{\sqrt{h^2+k^2}}\longrightarrow0.

Continuous first partial derivatives in a neighbourhood are sufficient. Differentiability implies continuity and coordinate partials; partials alone do not imply differentiability. The linear-approximation reference develops this distinction.

The remainder condition controls every small input increment, not just one selected path. It is the basis for treating the differential as a useful combined first-order prediction.

Compare a differential with an exact finite change

Take $g=x^2+3y^2$ at $(1,2)$ , with $\Delta x=0.10$ and $\Delta y=-0.05$ . The initial value is thirteen and the partials are $g_x=2$ , $g_y=12$ .

The differential is

dg=2(0.10)+12(-0.05)=-0.40.

Direct substitution instead gives

g(1.10,1.95)=1.21+3(3.8025)=12.6175,

\Delta g=12.6175-13=-0.3825.

Expansion identifies the difference exactly:

\Delta g=dg+(\Delta x)^2+3(\Delta y)^2,

\Delta g-dg=0.01+3(0.0025)=0.0175.

This is not a rounding accident. The first-order estimate omits quadratic terms. Their smallness makes the approximation useful, but reporting $-0.40$ as the exact change would discard a known part of the calculation.

Existing partials need not give continuity

Define $p(0,0)=0$ and $p(x,y)=xy/(x^2+y^2)$ elsewhere. Both coordinate traces through the origin are zero, so $p_x(0,0)=p_y(0,0)=0$ .

Along $y=x$ , the value is $1/2$ . The function is not continuous at the origin and therefore is not differentiable there. Writing a formal differential of zero does not rescue a tangent approximation whose hypothesis is absent.

The existence of both partial derivatives alone does not establish differentiability or justify a tangent-plane approximation.

Calculate higher and mixed partials

For $f=x^2y+3xy^2$ ,

f_{xx}=2y,\qquad f_{yy}=6x,

f_{xy}=2x+6y,\qquad f_{yx}=2x+6y.

Here $f_{xy}$ means first differentiating in $x$ , then differentiating that result in $y$ . Stating the convention avoids confusion with operator notation read from right to left.

Equality of mixed partials is guaranteed under suitable continuity of the second partial derivatives near the point. Polynomials satisfy those conditions. The partial-derivative reference gives the standard theorem.

Without those hypotheses, order can matter

Define $H(0,0)=0$ and

H(x,y)=\frac{xy(x^2-y^2)}{x^2+y^2}

elsewhere. Direct calculations on the axes give $H_x(0,y)=-y$ and $H_y(x,0)=x$ , including their zero values at the origin. Consequently,

H_{xy}(0,0)=-1,\qquad H_{yx}(0,0)=1.

This does not contradict the theorem because its continuity hypotheses fail. Agreement in many familiar examples is not a general licence to interchange derivative order.

Apply the chain rule when both inputs change

Along $x=t$ , $y=t^2$ , the polynomial becomes $f=t^4+3t^5$ . Its derivative is $4t^3+15t^4$ .

The multivariable chain rule gives

\frac{df}{dt}=f_xx’+f_yy’.

Here $f_x=2t^3+3t^4$ , $f_y=t^2+6t^3$ , $x’=1$ and $y’=2t$ . Combining them gives the same result. Using only $f_x$ would describe a different situation in which the second input remains fixed.

Use units to interpret sensitivity

If output is litres, $x$ is minutes and $y$ is metres, then $f_x$ is in litres per minute and $f_y$ in litres per metre. The products $f_x\Delta x$ and $f_y\Delta y$ both have litres as units and can be added.

The raw numerical sizes of partials with different units do not establish which input matters more. Compare changes on a justified scale. Derivatives describe local sensitivity; meaningful input increments make that sensitivity interpretable.

Practice questions

For $f=x^2y+3xy^2$ , find the partials at $(1,2)$ .
For the $g$ example, state $dg$ , $\Delta g$ and $\Delta g-dg$ .
Does the existence of two coordinate partial derivatives guarantee differentiability?

Check your answers

The partials are $2xy+3y^2$ and $x^2+6xy$ , giving sixteen and thirteen.
They are $-0.40$ , $-0.3825$ and $0.0175$ , respectively. The last quantity is the exact quadratic remainder.
No. The extended function $xy/(x^2+y^2)$ has both partials at the origin but is not continuous there. Continuity of the first partials in a neighbourhood is a sufficient stronger condition.