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Physics LibreTexts

2.2: Partial Derivatives

  • Page ID
    7215
  • The equation

    \[ z = z (x,~ y)\]

    represents a two-dimensional surface in three-dimensional space. The surface intersects the plane y = constant in a plane curve in which z is a function of x. One can then easily imagine calculating the slope or gradient of this curve in the plane y = constant. This slope is \( \left( \frac{\partial z}{\partial x} \right)_y\) - the partial derivative of z with respect to x, with y being held constant. For example, if

    \[ z = y \ln x,\]

    then

    \[ \left( \frac{\partial z}{\partial x} \right)_y = \frac{y}{x},\]

    y being treated as though it were a constant, which, in the plane y = constant, it is. In a similar manner the partial derivative of z with respect to y, with x being held constant, is

    \[ \left( \frac{\partial z}{\partial y} \right)_x = \ln x\]

    When you have only three variables – as in this example – it is usually obvious which of them is being held constant. Thus ∂z/∂y can hardly mean anything other than at constant x. For that reason, the subscript is often omitted. In thermodynamics, there are often more than three variables, and it is usually (I would say always) essential to indicate by a subscript which quantities are being held constant.

    In the matter of pronunciation, various attempts are sometimes made to give a special pronunciation to the symbol ∂. (I have heard “day”, and “dye”.) My own preference is just to say “partial dz by dy”.

    Let us suppose that we have evaluated z at (x , y). Now if you increase x by δx, what will the resulting increase in z be? Obviously, to first order, it is \( \frac{\partial x}{\partial x} \delta x\). And if y increases by δy, the increase in z will be \( \frac{\partial z}{\partial y} \delta y\). And if both x and y increase, the corresponding increase in z, to first order, will be

    \[ \delta z = \frac{\partial z}{\partial x} \delta x + \frac{\partial z}{\partial y} \delta y\]

    No great and difficult mathematical proof is needed to “derive” this; it is just a plain English statement of an obvious truism. The increase in z is equal to the rate of increase of z with respect to x times the increase in x plus the rate of increase of z with respect to y times the increase in y.

    Likewise if x and y are increasing with time at rates \( \frac{dx}{dt}\) and \( \frac{dy}{dt}\), the rate of increase of z with respect to time is

    \[ \frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt}.\]

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