Derivatives

2.1. Derivatives#

A function \(f(x,y)\) has partial derivatives:

(2.1)#\[\begin{equation} \td{}{f}{x}{y} = \lim_{h \to 0} \frac{f(x+h, y) - f(x, y)}{h} \end{equation}\]

where x (in the denominator) indicates what is varied and y (outside the parentheses) indicates what is held constant.

The second partial derivative may be “repeated”

(2.2)#\[\begin{equation} \td{2}{f}{x}{y} = \pp{}{}{x}\left[ \td{}{f}{x}{y} \right]_y \end{equation}\]

or “mixed”

(2.3)#\[\begin{equation} \frac{\partial^2 f}{\partial x \partial y} = \pp{}{}{x}\left[ \td{}{f}{y}{x} \right]_y \end{equation}\]

Mixed partial derivatives are read from right to left by convention.

Partial derivatives can be taken using the normal procedures for single-variable calculus if you treat the constant variables as such

Example: Taking partial derivatives

For the function \(f(x,y) = x^2 \cos y\), evaluate \((\partial f/\partial x)_y\), \((\partial^2 f/\partial x^2)_y\), and \(\partial^2 f/\partial y \partial x\).

The first partial derivative with respect to x (treating y as a constant) is

(2.4)#\[\begin{equation} \td{}{f}{x}{y} = 2 x \cos y \end{equation}\]

The second (repeated) partial derivative with respect to x (again, treating y as a constant) is

(2.5)#\[\begin{equation} \td{2}{f}{x}{y} = \pp{}{}{x}\left[ \td{}{f}{x}{y} \right]_y = \pp{}{}{x}\left( 2 x \cos y \right)_y = 2 \cos y \end{equation}\]

The second (mixed) partial derivative with respect to x then y is

(2.6)#\[\begin{equation} \frac{\partial^2 f}{\partial y \partial x} = \pp{}{}{y}\left[ \td{}{f}{x}{y} \right]_x = \pp{}{}{y} \left( 2 x \cos y \right)_x = -2x \sin y \end{equation}\]

Example: Order of mixed second partial derivatives

Show the order of the mixed derivatives does not matter for the example function given above.

The first partial derivative with respect to \(y\) is

(2.8)#\[\begin{equation} \td{}{f}{x}{y} = -x^2 \sin y \end{equation}\]

So, the second (mixed) partial derivative with respect to y then x is

(2.9)#\[\begin{equation} \frac{\partial^2 f}{\partial x \partial y} = \pp{}{}{x}\left[ \td{}{f}{y}{x} \right]_y = \pp{}{}{x} \left( -x^2 \sin y \right)_y = -2x \sin y \end{equation}\]

2.1.1. Skill builder problems#

Given:

(2.10)#\[\begin{align} f &= \cos(4x+y^2) + x^2y\\ x &= 2ut\\ y &= t^2 \end{align}\]

Evaluate:

  1. \(\displaystyle\td{}{f}{x}{y}\)

    Solution
    (2.11)#\[\begin{equation} \td{}{f}{x}{y} = -4\sin(4x+y^2) + 2xy \end{equation}\]

  2. \(\displaystyle\td{}{f}{y}{x}\)

    Solution
    (2.12)#\[\begin{equation} \td{}{f}{y}{x} = -2y\sin(4x+y^2) + x^2\\ \end{equation}\]

  3. \(\displaystyle\td{2}{f}{x}{y}\)

    Solution
    (2.13)#\[\begin{equation} \td{2}{f}{x}{y} = -16\cos(4x+y^2) + 2y \end{equation}\]

    (Differentiate #1.)

  4. \(\displaystyle\td{2}{f}{y}{x}\)

    Solution
    (2.14)#\[\begin{equation} \td{2}{f}{y}{x} = -4y^2\cos(4x+y^2) - 2\sin(4x+y^2) \end{equation}\]

    (Differentiate #1.)

  5. \(\displaystyle\frac{\partial^2 f}{\partial x \partial y}\)

    Solution
    (2.15)#\[\begin{equation} \frac{\partial^2 f}{\partial x \partial y} = -8y\cos(4x+y^2) + 2x \end{equation}\]

    (Differentiate #2.)

  6. \(\displaystyle\frac{\partial^2 f}{\partial y \partial x}\)

    Solution
    (2.16)#\[\begin{equation} \frac{\partial^2 f}{\partial y \partial x} = -8y\cos(4x+y^2) + 2x \end{equation}\]

    (Same as #5.)

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