2.1 Derivatives
A function f ( x , y ) f(x,y) f ( x , y ) partial derivatives:
( ∂ f ∂ x ) y = lim h → 0 f ( x + h , y ) − f ( x , y ) h \td{}{f}{x}{y} = \lim_{h \to 0} \frac{f(x+h, y) - f(x, y)}{h} ( ∂ x ∂ f ) y = h → 0 lim h f ( x + h , y ) − f ( x , y ) 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 f ∂ x 2 ) y = ∂ ∂ x [ ( ∂ f ∂ x ) y ] y \td{2}{f}{x}{y} = \pp{}{}{x}\left[ \td{}{f}{x}{y} \right]_y ( ∂ x 2 ∂ 2 f ) y = ∂ x ∂ [ ( ∂ x ∂ f ) y ] y or “mixed”
∂ 2 f ∂ x ∂ y = ∂ ∂ x [ ( ∂ f ∂ y ) x ] y \frac{\partial^2 f}{\partial x \partial y} = \pp{}{}{x}\left[ \td{}{f}{y}{x} \right]_y ∂ x ∂ y ∂ 2 f = ∂ x ∂ [ ( ∂ y ∂ f ) x ] y 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. For example, for the
function f ( x , y ) = x 2 cos y f(x,y) = x^2 \cos y f ( x , y ) = x 2 cos y
The first partial derivative with respect to x (treating y as a constant)
is
( ∂ f ∂ x ) y = 2 x cos y \td{}{f}{x}{y} = 2 x \cos y ( ∂ x ∂ f ) y = 2 x cos y The second (repeated) partial derivative with respect to x (again, treating
y as a constant) is
( ∂ 2 f ∂ x 2 ) y = ∂ ∂ x [ ( ∂ f ∂ x ) y ] y = ∂ ∂ x ( 2 x cos y ) y = 2 cos y \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 ( ∂ x 2 ∂ 2 f ) y = ∂ x ∂ [ ( ∂ x ∂ f ) y ] y = ∂ x ∂ ( 2 x cos y ) y = 2 cos y The second (mixed) partial derivative with respect to x then y is
∂ 2 f ∂ y ∂ x = ∂ ∂ y [ ( ∂ f ∂ x ) y ] x = ∂ ∂ y ( 2 x cos y ) x = − 2 x sin y \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 ∂ y ∂ x ∂ 2 f = ∂ y ∂ [ ( ∂ x ∂ f ) y ] x = ∂ y ∂ ( 2 x cos y ) x = − 2 x sin y Mixed second partial derivatives
If all the second partial derivatives of a multivariable function f are
continuous, the mixed second partial derivatives are symmetric,
∂ 2 f ∂ x ∂ y = ∂ 2 f ∂ y ∂ x \frac{\partial^2 f}{\partial x \partial y}
= \frac{\partial^2 f}{\partial y \partial x} ∂ x ∂ y ∂ 2 f = ∂ y ∂ x ∂ 2 f For the example function given above, the first partial derivative with respect
to y y y
( ∂ f ∂ x ) y = − x 2 sin y \td{}{f}{x}{y} = -x^2 \sin y ( ∂ x ∂ f ) y = − x 2 sin y So, the second (mixed) partial derivative with respect to y then x is
∂ 2 f ∂ x ∂ y = ∂ ∂ x [ ( ∂ f ∂ y ) x ] y = ∂ ∂ x ( − x 2 sin y ) y = − 2 x sin y \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 ∂ x ∂ y ∂ 2 f = ∂ x ∂ [ ( ∂ y ∂ f ) x ] y = ∂ x ∂ ( − x 2 sin y ) y = − 2 x sin y Skill builder problems ¶ Given:
f = cos ( 4 x + y 2 ) + x 2 y x = 2 u t y = t 2 \begin{align}
f &= \cos(4x+y^2) + x^2y\\
x &= 2ut\\
y &= t^2
\end{align} f x y = cos ( 4 x + y 2 ) + x 2 y = 2 u t = t 2 Evaluate:
( ∂ f ∂ x ) y = − 4 sin ( 4 x + y 2 ) + 2 x y \td{}{f}{x}{y} = -4\sin(4x+y^2) + 2xy ( ∂ x ∂ f ) y = − 4 sin ( 4 x + y 2 ) + 2 x y ( ∂ f ∂ y ) x = − 2 y sin ( 4 x + y 2 ) + x 2 \td{}{f}{y}{x} = -2y\sin(4x+y^2) + x^2\\ ( ∂ y ∂ f ) x = − 2 y sin ( 4 x + y 2 ) + x 2 ( ∂ 2 f ∂ x 2 ) y = − 16 cos ( 4 x + y 2 ) + 2 y \td{2}{f}{x}{y} = -16\cos(4x+y^2) + 2y ( ∂ x 2 ∂ 2 f ) y = − 16 cos ( 4 x + y 2 ) + 2 y (Differentiate #1.)
( ∂ 2 f ∂ y 2 ) x = − 4 y 2 cos ( 4 x + y 2 ) − 2 sin ( 4 x + y 2 ) \td{2}{f}{y}{x} = -4y^2\cos(4x+y^2) - 2\sin(4x+y^2) ( ∂ y 2 ∂ 2 f ) x = − 4 y 2 cos ( 4 x + y 2 ) − 2 sin ( 4 x + y 2 ) (Differentiate #1.)
∂ 2 f ∂ x ∂ y = − 8 y cos ( 4 x + y 2 ) + 2 x \frac{\partial^2 f}{\partial x \partial y} = -8y\cos(4x+y^2) + 2x ∂ x ∂ y ∂ 2 f = − 8 y cos ( 4 x + y 2 ) + 2 x (Differentiate #2.)
∂ 2 f ∂ y ∂ x = − 8 y cos ( 4 x + y 2 ) + 2 x \frac{\partial^2 f}{\partial y \partial x} = -8y\cos(4x+y^2) + 2x ∂ y ∂ x ∂ 2 f = − 8 y cos ( 4 x + y 2 ) + 2 x (Same as #5.)