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*
- Back to Integral Calculus
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