Partial Derivatives
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Function f(x,y) of two variables, there are two partial derivatives ∂f/∂x and ∂f/∂y.
 ∂f/∂x (a,b) = lim(h→0) f( a+h , b )  f( a , b )






 h






 ∂f/∂y (a,b) = lim(h→0) f( a , b+k )  f( a , b )






 k








 For ∂f/∂x , f is differentiated as a function with respect to x and y is treated as a constant.
 For ∂f/∂x , f is differentiated as a function with respect to y and x is treated as a constant.

Example 1 :
 Solve for ∂f/∂x and ∂f/∂y if f(x,y) = .
 Solution :
 ∂f/∂x = 2x
 ∂f/∂y = 1
Example 2 :
 Solve for ∂g/∂x (3,2) and ∂g/∂y (3,2) if g(x,y) = xye^.
 Solution :
 ∂g/∂x = (xy)*(2xe^) + ye^

 = 2ye^ + ye^ =ye^(2+1)

 ∂g/∂x (3,2) = 2e^(2+1)

 = 2(19)
 = 38

 ∂g/∂x = (xy)*(2xe^) + ye^

 ∂g/∂y = xe^
 ∂g/∂y (3,2) = 3
Example 3 :
 Solve for ∂f/∂x and ∂f/∂y if f(x,y)= ln(xy) +sin(x) = ln(x) + ln(y)+ sin(x).
 Solution :
 ∂f/∂x =
 ∂f/∂y =
Example 4 :
 Find the rectangular box of least surface area that has volume 1000.
 Solution :
 Volume = 1000 = xyz
 Surface Area = 2xz + 2yz + 2xy

 z = 1000/xy
 S = 2x(1000/xy) + 2y(1000/xy) + 2xy
 = 2000/y + 2000/x + 2xy *need to minimize this*

 Solve for ∂S/∂x = 0 and ∂S/∂y = 0
 ∂S/∂x = 2000/ + 2y = 0
 ∂S/∂y = 2000/ + 2x = 0
 y = 1000/ and x = 1000/
 y() = 0
 Therefore, y = 0 or 1000 = 0
 So, y = 10 and x = 10.
 Using the second derivative test, S does have a local minimum at (10, 10).
 Therefore, y = 0 or 1000 = 0
 y() = 0
 y = 1000/ and x = 1000/
 Solve for ∂S/∂x = 0 and ∂S/∂y = 0

 = 2000/y + 2000/x + 2xy *need to minimize this*

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