Functions of several variables — it's simpler than you think.
A molecule's energy depends on multiple coordinates:
$$V(r_1, r_2, \theta) = \text{function of bond lengths and angle}$$
For water: V depends on two O-H distances and the H-O-H angle.
Question: How does V change if we stretch just one bond, holding everything else fixed?
Answer: That's a partial derivative.
$$\frac{\partial V}{\partial r_1} = \text{rate of change of } V \text{ with respect to } r_1 \text{, holding } r_2 \text{ and } \theta \text{ constant}$$
Partial differentiation = ordinary differentiation with some variables treated as constants.
That's the whole idea. Everything else is notation.
| Situation | What You Need |
|---|---|
| "How does Y change with X, holding Z fixed?" | ∂Y/∂X |
| Energy depends on multiple coordinates | ∂V/∂qi |
| Thermodynamic state functions | (∂G/∂T)P, (∂G/∂P)T |
| Finding equilibrium geometry | Set all ∂V/∂qi = 0 |
| Checking stability | Hessian matrix of second partials |
To find ∂f/∂x:
That's it.
∂f/∂x (treat y as constant):
$$\frac{\partial f}{\partial x} = 2xy + 3y^2 - 0 = 2xy + 3y^2$$
(The −y³ term has no x, so its derivative w.r.t. x is 0.)
∂f/∂y (treat x as constant):
$$\frac{\partial f}{\partial y} = x^2 + 6xy - 3y^2$$
Using the chain rule with y as constant:
$$\frac{\partial f}{\partial x} = ye^{xy}, \quad \frac{\partial f}{\partial y} = xe^{xy}$$
| Notation | Meaning |
|---|---|
| ∂f/∂x | Partial of f with respect to x |
| fx | Same (subscript notation) |
| (∂f/∂x)y | Explicitly showing y is held constant |
| fx(a, b) | Partial evaluated at point (a, b) |
The subscript outside the parentheses shows what's held constant. This is essential in thermodynamics.
Gradient: The vector of all partial derivatives:
$$\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$$
The gradient points in the direction of steepest ascent. Its magnitude is the rate of steepest ascent.
For potential energy: Force = −∇V (points downhill)
Critical point: Where ∇f = 0 (all partial derivatives are zero).
At a critical point (a, b), compute:
$$D = f_{xx} \cdot f_{yy} - (f_{xy})^2 = \det(\text{Hessian})$$
| Condition | Type |
|---|---|
| D > 0 and fxx > 0 | Local minimum |
| D > 0 and fxx < 0 | Local maximum |
| D < 0 | Saddle point |
| D = 0 | Inconclusive |
Hessian: The matrix of second partial derivatives:
$$\mathbf{H} = \begin{pmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{pmatrix}$$
Eigenvalues of H determine the nature of critical points:
If fxy and fyx are both continuous, then:
$$f_{xy} = f_{yx}$$
The order of differentiation doesn't matter.
P as a function of V and T: $P = \frac{nRT}{V}$
$$\left(\frac{\partial P}{\partial T}\right)_V = \frac{nR}{V}$$
(At constant volume, pressure increases linearly with temperature.)
$$\left(\frac{\partial P}{\partial V}\right)_T = -\frac{nRT}{V^2}$$
(At constant temperature, pressure decreases as volume increases.)
For state functions:
$$\left(\frac{\partial P}{\partial T}\right)_V \left(\frac{\partial T}{\partial V}\right)_P \left(\frac{\partial V}{\partial P}\right)_T = -1$$
A molecule's equilibrium geometry minimizes V(q1, q2, ..., qn):
Necessary condition: $\frac{\partial V}{\partial q_i} = 0$ for all i (all forces zero)
Hessian eigenvalues give vibrational frequencies:
Total differential: How f changes when x and y both change:
$$df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy$$
If f depends on measured quantities x and y with uncertainties δx and δy:
$$(\delta f)^2 \approx \left(\frac{\partial f}{\partial x}\right)^2(\delta x)^2 + \left(\frac{\partial f}{\partial y}\right)^2(\delta y)^2$$
Example: V = πr²h with r = 2.0 ± 0.1 cm, h = 5.0 ± 0.2 cm gives V = 62.8 ± 6.8 cm³
| Concept | Key Formula |
|---|---|
| ∂f/∂x | Differentiate treating other variables as constants |
| Gradient | ∇f = (∂f/∂x, ∂f/∂y, ...) |
| Force from potential | F = −∇V |
| Critical points | Where ∇f = 0 |
| Hessian | Hij = ∂²f/∂xi∂xj |
| Total differential | df = ∑ (∂f/∂xi)dxi |
| Mixed partials | fxy = fyx |