91Ó°ÊÓ

We are given the equation \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\). We can solve for z by subtracting the terms \(\frac{1}{x}\) and \(\frac{1}{y}\) from both sides: $$\frac{1}{z} = 1 - \frac{1}{x} - \frac{1}{y}$$ Now, take the reciprocal of both sides to find an expression for z: $$z = \frac{1}{1 - \frac{1}{x} - \frac{1}{y}}$$

Now we will find the partial derivative of z with respect to x. First, note that we can rewrite the denominator using a common denominator: $$z = \frac{1}{\frac{x\cdot y}{x\cdot y} - \frac{y}{x\cdot y} - \frac{x}{x \cdot y}} = \frac{1}{\frac{xy - x - y}{xy}}$$ So, z can be written as: $$z = \frac{xy}{xy - x - y}$$ Now we can find the partial derivative with respect to x using the quotient rule: $$\frac{\partial z}{\partial x}=\frac{(xy - x - y)\left(\frac{\partial (xy)}{\partial x}\right) - (xy)\left(\frac{\partial (xy-x-y)}{\partial x}\right)}{(xy - x - y)^2}$$ Compute the partial derivatives in the numerator: $$\frac{\partial (xy)}{\partial x} = y \hspace{10mm} \text{and} \hspace{10mm} \frac{\partial (xy-x-y)}{\partial x} = y - 1$$ Substitute these values into the partial derivative formula: $$\frac{\partial z}{\partial x}=\frac{(xy - x - y)(y) - (xy)(y - 1)}{(xy - x - y)^2}$$ Now, simplify the expression for the partial derivative: $$\frac{\partial z}{\partial x}=\frac{xy^2 - xy + y^2 - x^2y + x^2 + xy - y^2}{(xy - x - y)^2}$$

Cancel out the terms in the numerator: $$\frac{\partial z}{\partial x}=\frac{x^2 - x^2y + xy}{(xy - x - y)^2}$$ And there you have your the partial derivative of z with respect to x: $$\frac{\partial z}{\partial x}=\frac{x^2 - x^2y + xy}{(xy - x - y)^2}$$