SOLUTION: 2. Suppose y=100-5x and z=x(y-1) Δy/Δx=? ∂z/∂x=? ∂z/∂y=? Δz/Δx=?

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y = 100 - 5x

Substitute (y+Δy) for y and (x+Δx) for x

  y+Δy = 100 - 5(x+Δx)

y + Δy = 100 - 5x + 5Δx

    Δy = 100 - 5x + 5Δx - y     

Now substitute (100-5x) for y

    Δy = 100 - 5x + 5Δx - (100-5x)

    Δy = 100 - 5x + 5Δx - 100 + 5x

    Δy = -5Δx  

 Δy/Δx = -5Δx/Δx

 Δy/Δx = -5   


z = x(y - 1)

To find ∂z/∂x, we hold y constant, which means that
(y - 1) is a constant, and

∂z/∂x = y - 1

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z = x(y - 1)

To find ∂z/∂y, we hold x constant, which means that
x is a constant, and so

∂z/∂y = x

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     z = x(y - 1)

Substitute (z+Δz) for z, (y+Δy) for y, and (x+Δx) for x

    z+Δz = (x+Δx)[(y+Δy) - 1]

  z + Δz = (x + Δx)(y + Δy - 1)

  z + Δz = xy + xΔy - x + yΔx + ΔxΔy - Δx
      
      Δz = xy + xΔy - x + yΔx + ΔxΔy - Δx - z

Substitute  x(y - 1) for z

      Δz = xy + xΔy - x + yΔx + ΔxΔy - Δx - x(y - 1)

      Δz = xy + xΔy - x + yΔx + ΔxΔy - Δx - xy + x

      Δz = xΔy + yΔx + ΔxΔy - Δx

   Δz/Δx = xΔy/Δx + y + Δy - 1

Now substitute 100 - 5x for y, -5Δx for Δy, and -5 for Δy/Δx

   Δz/Δx = x(-5) + (100 - 5x) + (-5Δx) - 1

   Δz/Δx = -5x + 100 - 5x - 5Δx - 1

   Δz/Δx = -10x + 99 - 5x - 5Δx

Edwin