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Found in: Page 657

### Essential Calculus: Early Transcendentals

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.

See the step by step solution

## Step 1: Formula used

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$$F(x) = z - f(x - y)$$.

## Step 2: Determine the partial derivatives of $$\frac{{\partial z}}{{\partial x}}$$and $$\frac{{\partial z}}{{\partial y}}$$

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.