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Q39E

Expert-verifiedFound in: Page 657

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

**Show the equation \(\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}} = 0\) holds true.**

The equation \(\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}} = 0\) holds true.

**Equation 7: " **\(\frac{{\partial z}}{{\partial x}} = - \frac{{\frac{{\partial F}}{{dx}}}}{{\frac{{\partial F}}{{dz}}}}\)** and **\(\frac{{\partial z}}{{\partial y}} = - \frac{{\frac{{\partial F}}{{\partial y}}}}{{\frac{{\partial F}}{{\partial x}}}}\)

**The function **\(F\)** becomes**\(F(x) = z - f(x - y)\)**.**

As given that the function is \(z = f(x - y)\)

That is, \(z - f(x - y) = 0\)

Obtain the value of \(\frac{{\partial z}}{{\partial x}}\) as follows

\(\begin{aligned}{l}\frac{{\partial z}}{{\partial x}} = - \frac{{\frac{\partial }{{\partial x}}(z - f(x - y))}}{{\frac{\partial }{\partial }(z - f(x - y))}}\\\frac{{\partial z}}{{\partial x}} = \frac{{ - {f^\prime }(x - y)}}{1}\\\frac{{\partial z}}{{\partial x}} = - {f^\prime }(x - y)\end{aligned}\)

Obtain the value of \(\frac{{\partial z}}{{\partial y}}\) as follows,

\(\begin{aligned}{l}\frac{{\partial z}}{{\partial y}} = - \frac{{\frac{\partial }{{\partial y}}(z - f(x - y))}}{{\frac{\partial }{\partial }(z - f(x - y))}}\\\frac{{\partial z}}{{\partial y}} = \frac{{ - {f^\prime }(x - y)( - 1)}}{1}\\\frac{{\partial z}}{{\partial y}} = {f^\prime }(x - y)\end{aligned}\)

Thus, the sum of the partial derivatives becomes\(\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}} = - {f^\prime }(x - y) + {f^\prime }(x - y) = 0\). Hence, the equation \(\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}} = 0\) holds true.

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