Partial Differentiation
Partial Derivatives
The volume of a circular cylinder of radius r and height h is
V = πr2 h;
we say that V is a function of the two independent variables r and h. Thus
V = f (r, h)
where f (r, h) = π 2 rh (r > 0, h > 0). If h is held constant then we can differentiate f with respect to
r. This is the partial derivative of f with respect to r and is denoted by
∂f
or fr or f1 ,
∂r
where 1 denotes the first independent variable r. It follows that
∂f
∂
=
πr2 h
∂r
∂r
∂
r2
(since π and h are constant)
= πh
∂r
= πh2r
= 2πrh.
Similarly, if r is held constant, we can differentiate f with respect to h. This is the partial derivative
of f with respect to h and is denoted by
∂f
or fh or f2 ,
∂h
where 2 denotes the second independent variable h. Therefore
∂f
∂
=
πr2 h
∂h
∂h
∂
(h) (since π and r are constant)
= πr2
∂h
= πr2 .
Partial derivatives of functions of several variables are calculated in the same way.
Example 1 Let f (x, y, z) = x3 y 4 z 5 + x + y 2 + z 3 . Then
∂f
= 3x2 y 4 z 5 + 1,
∂x
∂f
= 4x3 y 3 z 5 + 2y,
∂y
∂f
= 5x3 y 4 z 4 + 3z 2 .
∂z
The chain, product and quotient rules for derivatives of one variable extend naturally to partial
derivatives.
Example 2 (Chain Rule) Let f (x, y) = sin(x2 y 3 + y 2 ). Then
∂f
∂x
∂
sin(x2 y 3 + y 2 )
∂x
∂
= cos(x2 y 3 + y 2 )
x2 y 3 + y 2
∂x
2 3
2
= cos(x y + y )(2xy 3 ).
=
1
(by the chain rule)
Exercise 1 Show that
∂f
= (3x2 y 2 + 2y) cos(x2 y 3 + y 2 ).
∂y
2 y2
Example 3 (Product Rule) Let f (x, y) = (x2 + y 2 )ex
∂f
∂x
. Then
∂
∂ x2 y 2 2 2
x2 + y 2 · ex y + (x2 + y 2 ) ·
e
(by the product rule)
∂x
∂x
2 2
2 2 ∂
= 2xex y + (x2 + y 2 )ex y
x2 y 2
(by the chain rule)
∂x
2 2
2 2
= 2xex y + 2xy 2 (x2 + y 2 )ex y
=
= 2xex
2 y2
(1 + x2 y 2 + y 4 ).
Exercise 2 Show that
∂f
2 2
= 2yex y (1 + x2 y 2 + x4 ).
∂y
Example 4 (Quotient Rule) Let f (x, y, z) =
∂f
∂z
=
=
=
=
∂
∂z
xz
y+z
∂
∂z
xz
. Then
y+z
(xz) · (y + z) − xz ·
(y + z)2
x(y + z) − xz
(y + z)2
xy
.
(y + z)2
Exercise 3 Show that
and
∂
∂z
(y + z)
(by the quotient rule)
∂f
z
=
∂x
y+z
∂f
xz
=−
.
∂y
(y + z)2
Higher Partial Derivatives
As with ordinary derivatives of functions of one variable, we can compute higher partial derivatives of
functions of several variables. If f is a function of x and y then
∂2f
or fxx or f11
∂x2
denote the second partial derivative of f with respect to x. Similarly,
∂2f
or fyy or f22
∂y 2
denote the second partial derivative of f with respect to y. The notation
∂2f
or fxy or f12
∂x∂y
is used to denote the partial derivative of f first with respect to x and then with respect to y. If f is
a suitably well-behaved function then fxy = fyx , where fyx has its obvious meaning.
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Example 5 Let f (x, y) = x3 y 2 + x−1 + y −1 . Then
∂
∂
3 2
−1
−1
fxx =
x y +x +y
∂x ∂x
∂
3x2 y 2 − x−2
=
∂x
= 6xy 2 + 2x−3 ,
fyy
fxy
fyx
∂
∂
3 2
−1
−1
=
x y +x +y
∂y ∂y
∂
=
2x3 y − y −2
∂y
= 2x3 + 2y −3 ,
∂
∂
3 2
−1
−1
x y +x +y
=
∂y ∂x
∂
=
3x2 y 2 − x−2
∂y
= 6x2 y,
∂
∂
3 2
−1
−1
x y +x +y
=
∂x ∂y
∂
2x3 y − y −2
=
∂x
= 6x2 y.
Implicit Partial Differentiation
As with functions of one variable, we can compute the partial derivatives of a function defined implicitly
in terms of several variables. For example, the function f is defined implicitly in terms of x and y by
the equation
sin(yf ) = x + y,
and its partial x-derivative is computed as follows:
∂
∂
(sin(yf )) =
(x + y)
∂x
∂x
∂
⇒ cos(yf )
(yf ) = 1 (by the chain rule)
∂x
⇒ yfx cos(yf ) = 1
1
.
⇒ fx =
y cos(yf )
Similarly
∂
∂
(sin(yf )) =
(x + y)
∂y
∂x
∂
⇒ cos(yf ) (yf ) = 1 (by the chain rule)
∂y
⇒ cos(yf )(f + yfy ) = 1 (by the product rule)
⇒ f cos(yf ) + yfy cos(yf ) = 1
1 − f cos(yf )
⇒ fy =
.
y cos(yf )
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