Question 124369
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+&#916;y) for y and (x+&#916;x) for x

  y+&#916;y = 100 - 5(x+&#916;x)

y + &#916;y = 100 - 5x + 5&#916;x

    &#916;y = 100 - 5x + 5&#916;x - y     

Now substitute (100-5x) for y

    &#916;y = 100 - 5x + 5&#916;x - (100-5x)

    &#916;y = 100 - 5x + 5&#916;x - 100 + 5x

    &#916;y = -5&#916;x  

 &#916;y/&#916;x = -5&#916;x/&#916;x

 &#916;y/&#916;x = -5   


z = x(y - 1)

To find &#8706;z/&#8706;x, we hold y constant, which means that
(y - 1) is a constant, and

&#8706;z/&#8706;x = y - 1

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z = x(y - 1)

To find &#8706;z/&#8706;y, we hold x constant, which means that
x is a constant, and so

&#8706;z/&#8706;y = x

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     z = x(y - 1)

Substitute (z+&#916;z) for z, (y+&#916;y) for y, and (x+&#916;x) for x

    z+&#916;z = (x+&#916;x)[(y+&#916;y) - 1]

  z + &#916;z = (x + &#916;x)(y + &#916;y - 1)

  z + &#916;z = xy + x&#916;y - x + y&#916;x + &#916;x&#916;y - &#916;x
      
      &#916;z = xy + x&#916;y - x + y&#916;x + &#916;x&#916;y - &#916;x - z

Substitute  x(y - 1) for z

      &#916;z = xy + x&#916;y - x + y&#916;x + &#916;x&#916;y - &#916;x - x(y - 1)

      &#916;z = xy + x&#916;y - x + y&#916;x + &#916;x&#916;y - &#916;x - xy + x

      &#916;z = x&#916;y + y&#916;x + &#916;x&#916;y - &#916;x

   &#916;z/&#916;x = x&#916;y/&#916;x + y + &#916;y - 1

Now substitute 100 - 5x for y, -5&#916;x for &#916;y, and -5 for &#916;y/&#916;x

   &#916;z/&#916;x = x(-5) + (100 - 5x) + (-5&#916;x) - 1

   &#916;z/&#916;x = -5x + 100 - 5x - 5&#916;x - 1

   &#916;z/&#916;x = -10x + 99 - 5x - 5&#916;x

Edwin</pre>