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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
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
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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