Partial Derivatives in Calculus

Let f(x,y) be a function with two variables. If we keep y constant and differentiate f (assuming f is differentiable) with respect to the variable x, we obtain what is called the partial derivative of f with respect to x which is denoted by
∂f
∂x
or fx
Similarly If we keep x constant and differentiate f (assuming f is differentiable) with respect to the variable y, we obtain what is called the partial derivative of f with respect to y which is denoted by
∂f
∂y
or fy

We might also define partial derivatives of function f as follows:


∂f
∂x
= lim
h→0
f(x + h , y) - f(x , y)
h
∂f
∂y
= lim
k→0
f(x , y + k) - f(x , y)
k

We now present several examples with detailed solution on how to calculate partial derivatives.

Example 1: Find the partial derivatives fx and fy if f(x , y) is given by

f(x , y) = x2 y + 2x + y

Solution to Example 1:

Assume y is constant and differentiate with respect to x to obtain

fx = ∂f
∂x
=
∂x
[ x2 y + 2x + y ]
=
∂x
[ x2 y] +
∂x
[ 2 x ] +
∂x
[ y ] = [2 x y] + [ 2 ] + [ 0 ] = 2x y + 2
Now assume x is constant and differentiate with respect to y to obtain

fy = ∂f
∂y
=
∂y
[ x2 y + 2x + y ]
=
∂y
[ x2 y] +
∂y
[ 2 x ] +
∂y
[ y ] = [ x2 ] + [ 0 ] + [ 1 ] = x2 + 1

Example 2: Find fx and fy if f(x , y) is given by

f(x , y) = sin(x y) + cos x

Solution to Example 2:

Differentiate with respect to x assuming y is constant

fx = ∂f
∂x
=
∂x
[ sin(x y) + cos x ] = y cos(x y) - sin x


Differentiate with respect to y assuming x is constant
fy = ∂f
∂y
=
∂y
[ sin(x y) + cos x ] = x cos(x y)


Example 3: Find fx and fy if f(x , y) is given by

f(x , y) = x ex y

Solution to Example 3:

Differentiate with respect to x assuming y is constant

fx = ∂f
∂x
=
∂x
[ x ex y ] = ex y + x y ex y = (x y + 1)ex y


Differentiate with respect to y
fy = ∂f
∂y
=
∂y
[ x ex y ] = (x) (x ex y) = x2 ex y

Example 4: Find fx and fy if f(x , y) is given by

f(x , y) = ln ( x2 + 2 y)

Solution to Example 4:

Differentiate with respect to x to obtain

fx = ∂f
∂x
=
∂x
[ ln ( x2 + 2 y) ] = 2x
x2 + 2 y


Differentiate with respect to y
fy = ∂f
∂y
=
∂y
[ ln ( x2 + 2 y) ] = 2
x2 + 2 y


Example 5: Find fx(2 , 3) and fy(2 , 3) if f(x , y) is given by

f(x , y) = y x2 + 2 y

Solution to Example 5:

We first find fx and fy

fx(x,y) = 2x y

fy(x,y) = x2 + 2

We now calculate fx(2 , 3) and fy(2 , 3) by substituting x and y by their given values

fx(2,3) = 2 (2)(3) = 12

fy(2,3) = 22 + 2 = 6


Exercise: Find partial derivatives fx and fy of the following functions

1. f(x , y) = x ex + y

2. f(x , y) = ln ( 2 x + y x)

3. f(x , y) = x sin(x - y)

Answer to Above Exercise:

1. fx =(x + 1)ex + y , fy = x ex + y

2. fx = 1 / x , fy = 1 / (y + 2)

3. fx = x cos (x - y) + sin (x - y) , fy = -x cos (x - y)

More on partial derivatives and mutlivariable functions. Multivariable Functions



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Updated: 2 April 2013

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