# Partial Derivatives

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) = ${\displaystyle -x^{2}+y}$.
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^${\displaystyle x^{2}}$.
Solution :
∂g/∂x = (xy)*(2xe^${\displaystyle x^{2}}$) + ye^${\displaystyle x^{2}}$
= 2${\displaystyle x^{2}}$ye^${\displaystyle x^{2}}$ + ye^${\displaystyle x^{2}}$ =ye^${\displaystyle x^{2}}$(2${\displaystyle x^{2}}$+1)
∂g/∂x (3,2) = 2e^${\displaystyle 3^{2}}$(2${\displaystyle 3^{2}}$+1)
= 2${\displaystyle e^{9}}$(19)
= 38${\displaystyle e^{9}}$

∂g/∂y = xe^${\displaystyle x^{2}}$
∂g/∂y (3,2) = 3${\displaystyle e^{9}}$

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 = ${\displaystyle 1/xy*y+cos(x)=1/x+cos(x)}$
∂f/∂y = ${\displaystyle 1/xy*x=1/y}$

Example 4 :

Find the rectangular box of least surface area that has volume 1000${\displaystyle in^{3}}$.
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/${\displaystyle x^{2}}$ + 2y = 0
∂S/∂y = -2000/${\displaystyle y^{2}}$ + 2x = 0
y = 1000/${\displaystyle x^{2}}$ and x = 1000/${\displaystyle y^{2}}$
y(${\displaystyle y^{3}-1000}$) = 0
Therefore, y = 0 or ${\displaystyle y^{3}}$-1000 = 0
So, y = 10 and x = 10.
Using the second derivative test, S does have a local minimum at (10, 10).