Problem 51 - Calculus Early Transcendentals

Question

Use the result of Exercise 48 to evaluate \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) for the following relations. $$x y z+x+y-z=0$$

Step-by-Step Solution

Answer

Question: Find the partial derivatives of \(z\) with respect to \(x\) and \(y\) given the relation \(xyz + x + y - z = 0\). Answer: The partial derivatives of \(z\) are: \(\frac{\partial z}{\partial x} = \frac{(xy - 1)(1) - (x + y)(y)}{(xy - 1)^2}\) \(\frac{\partial z}{\partial y} = \frac{(xy - 1)(1) - (x + y)(x)}{(xy - 1)^2}\)

Step 1: Isolate z

To isolate z, we need to rewrite the equation such that z is on one side of the equation and the rest of the variables are on the other side. The given relation is \(xyz + x + y - z = 0\). So, let's isolate z: \(z(x y - 1) = x + y\) Now divide both sides by $(xy - 1): \(z = \frac{x+y}{xy - 1}\) Now that we have z in terms of x and y, we can find the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\).

Step 2: Computing \(\frac{\partial z}{\partial x}\)

To compute the partial derivative of z with respect to x, we differentiate z with respect to x while treating y as a constant: \(\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}\left(\frac{x+y}{xy - 1}\right)\) To differentiate this fraction, we can use the quotient rule, which states that the derivative of \(\frac{u}{v}\) is \(\frac{vu' - uv'}{v^2}\), where u and v are functions of x and y, and u' and v' are their respective derivatives. Here, \(u = x + y\) and \(v = xy - 1\). To find their derivatives with respect to x, we have: \(u' = \frac{\partial}{\partial x}(x + y) = 1\) \(v' = \frac{\partial}{\partial x}(xy - 1) = y\) Now we apply the quotient rule: \(\frac{\partial z}{\partial x} = \frac{v u' - u v'}{v^2} = \frac{(xy - 1)(1) - (x + y)(y)}{(xy - 1)^2}\)

Step 3: Computing \(\frac{\partial z}{\partial y}\)

To compute the partial derivative of z with respect to y, we will differentiate z with respect to y while treating x as a constant: \(\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}\left(\frac{x+y}{xy - 1}\right)\) We already have expressions for u, v, and their derivatives with respect to x. Now we need to find their derivatives with respect to y: \(u' = \frac{\partial}{\partial y}(x + y) = 1\) \(v' = \frac{\partial}{\partial y}(xy - 1) = x\) Now applying the quotient rule: \(\frac{\partial z}{\partial y} = \frac{v u' - u v'}{v^2} = \frac{(xy - 1)(1) - (x + y)(x)}{(xy - 1)^2}\) So the partial derivatives of z are: \(\frac{\partial z}{\partial x} = \frac{(xy - 1)(1) - (x + y)(y)}{(xy - 1)^2}\) \(\frac{\partial z}{\partial y} = \frac{(xy - 1)(1) - (x + y)(x)}{(xy - 1)^2}\)